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Where a polynomial is zero
A STORY
Where the arch meets the ground
Ananya and Aarav stand by the river, looking up at the stone arch. The arch leaves the ground on
one bank. It rises high overhead. It comes back down to the ground on the far bank.
“It meets the ground twice,” Ananya says. “Once here, once on the far bank.”
Aarav points up at the arch, then across to the far bank. “That is $40$ m from here,” he says.
“I paced it out on the way here.”
“The engineer had a rule for the arch before it was built,” Ananya says. “A rule that gives both
distances, without walking either one.”
“Just from an equation?” Aarav asks.
“Just from an equation,” Ananya says. “Two numbers come straight out of it. They are exactly
where the arch meets the ground.”
What equation would give you both of those numbers, before you walk to either bank?
Take $p(x) = x - 2$. Put $x = 2$ into it. $p(2) = 2 - 2 = 0$. A number that sends a polynomial’s value to $0$ is called a ZERO.
Say the same rule with a letter. $k$ is a zero of $p(x)$ when $p(k) = 0$.
A zero is also a point on a graph. Draw $y = p(x)$ and you will see it. The zero is the point where the curve meets the x-axis. For $p(x) = x - 2$, that point is $(2, 0)$. Look at the picture and the zero sits right there. No working needed.
We use one fact for two jobs in this chapter.
Job one: read a zero off the graph, at the x-axis crossing.
Job two: find a zero from the equation, by working it out.
Both jobs land on the same numbers. For a quadratic $a x^2 + b x + c$, splitting the middle term finds both zeros. Their sum is $-b/a$. Their product is $c/a$. We reach both by the end of the chapter.
Your turn: is $x = 3$ a zero of $p(x) = x - 2$? (Answer: no. $p(3) = 3 - 2 = 1$, not $0$.)
Only the left crossing is worked out on the card, and the right one is marked and left for you to check.
A SCHOLAR INDIA REMEMBERS
Brahmagupta worked at Bhillamala, in today’s Rajasthan, around 628 CE. He was the first to write down rules for zero as a number in its own right. You can add zero, take it away, and multiply by it. For example, $5 + 0 = 5$ and $5 \times 0 = 0$. A zero of a polynomial is a value of $x$ that makes the polynomial equal to that number, $0$.
Brahmagupta · the brahminy kite
Check yourself
A zero of a polynomial $p(x)$ is a value $k$ such that
$k = 0$
$p(x)$ equals $k$ for every $x$
$p(k) = 0$
$k$ is the degree of $p(x)$
Check your answer
$k = 0$ — The zero is the value $k$ that makes $p(k)=0$ — $k$ itself need not be $0$.
$p(x)$ equals $k$ for every $x$ — $p(k)=0$ is a statement about one point, $x=k$, not about $p(x)$ for every $x$.
✓ $p(k) = 0$ — (C) A zero is defined by the value the polynomial takes at $k$, not by $k$ itself.
$k$ is the degree of $p(x)$ — The degree counts the highest power in $p(x)$; it has no connection to which value is a zero.
For $p(x) = x^2 - 5x + 6$, which value of $x$ is a zero?
$x = 2$, since $p(2) = 4 - 10 + 6 = 0$
$x = 6$, since $6$ is the constant term
$x = 5$, since $5$ is the coefficient of $x$
$x = 1$, since $p(1) = 1 - 5 + 6 = 2$ is close to zero
Check your answer
✓ $x = 2$, since $p(2) = 4 - 10 + 6 = 0$ — (A) Substituting $2$ for $x$ and simplifying gives exactly $0$, which is what a zero requires.
$x = 6$, since $6$ is the constant term — The constant term is not automatically a zero; substituting $x=6$ gives $p(6)=36-30+6=12 \neq 0$.
$x = 5$, since $5$ is the coefficient of $x$ — A coefficient is not a zero; substituting $x=5$ gives $p(5)=25-25+6=6 \neq 0$.
$x = 1$, since $p(1) = 1 - 5 + 6 = 2$ is close to zero — A zero must give exactly $0$; $p(1)=2$, and $2$ is not $0$, however small it looks.
On the graph of $y = p(x)$, a zero of $p(x)$ corresponds to
a point where the curve meets the x-axis
the highest point on the curve
the point where the curve crosses itself
the slope of the curve at $x = 0$
Check your answer
✓ a point where the curve meets the x-axis — (A) A zero is where the function value is $0$, and points with $y=0$ lie on the x-axis.
the highest point on the curve — The highest (or lowest) point of a curve is its vertex, not a zero — a zero is about crossing the x-axis, not turning.
the point where the curve crosses itself — A polynomial graph does not cross itself; a zero is simply a meeting with the x-axis.
the slope of the curve at $x = 0$ — Slope measures how steeply the curve rises or falls; a zero is a location where the curve meets the x-axis, not a rate.
You have used letters for numbers before. Now those letters make a new kind of expression. Try each check below. It takes a minute.
Degree of a polynomial (Class 9, Ganita Manjari, page 18). Here: a quadratic, like $x^2 + 5x + 1$, is named for reaching degree $2$. Check: $x^2 + 5x + 1$ has degree $2$, the highest power of $x$.
Substituting into a polynomial (Class 9, Ganita Manjari, page 20). Here: finding $p(k)$ means substituting $k$ for $x$, exactly as here. Check: $2 \cdot 4 + 3 = 11$.
A point on a line (Class 9, Ganita Manjari, page 28). Here: a zero of $p(x)$ is a point where its graph crosses the x-axis. Check: $(7, 15)$ satisfies $y = 2x + 1$, since $2 \cdot 7 + 1 = 15$.
Difference of two squares (Class 9, Ganita Manjari, page 77). Here: this identity turns $x^2 - 3$ into its two irrational zeros. Check: $5^2 - 2^2 = (5 - 2)(5 + 2) = 21$.
Factorising a quadratic (Class 9, Ganita Manjari, page 81). Here: every split starts from two numbers with this sum and product. Check: $a + b = 7$ and $a \cdot b = 12$ give $a = 3$, $b = 4$.
Expanding two brackets (Class 8, Ganita Prakash Part 1, page 140). Here: splitting the middle term ends in this bracket multiplication. Check: $(5 - 2)(3 + 4) = 5 \cdot 3 - 2 \cdot 3 + 5 \cdot 4 - 2 \cdot 4 = 21$.
If any of these felt new, read the page named before going on.
Every input gives a value, but only the inputs whose value is 0 earn the name zero.
Take $p(x) = x^2 - 3 x - 4$. Put $x = 2$ in place of $x$. $p(2) = 2^2 - 3 \cdot 2 - 4$. Work it out one piece at a time: $2^2 = 4$, $3 \cdot 2 = 6$, so $p(2) = 4 - 6 - 4 = -6$. That number, $-6$, is the VALUE of $p(x)$ at $x = 2$.
Say it again with a letter. The value of $p(x)$ at $x = k$ is written $p(k)$. Find it the same way we just did. Put $k$ in place of $x$ everywhere in $p(x)$, then work through the arithmetic in order.
Your turn: find $p(-1)$ for the same polynomial. (Answer: $p(-1) = 1 + 3 - 4 = 0$.)
*A zero is the special case where this value comes out to $0$: $p(k) = 0$ means $k$ is a zero of $p(x)$.*
Check yourself
For a polynomial $p(x)$, the value $p(3)$ means
multiply $p(x)$ by $3$
substitute $3$ for $x$
add $3$ to $p(x)$
solve $p(x) = 3$ for $x$
Check your answer
multiply $p(x)$ by $3$ — $p(3)$ is not $p(x)$ scaled by $3$ — it is $p(x)$ evaluated AT $x=3$.
✓ substitute $3$ for $x$ — (B) The value of a polynomial at a number is found by substitution, then simplifying.
add $3$ to $p(x)$ — $p(3)$ substitutes $3$ for $x$; it does not add $3$ to the whole expression.
solve $p(x) = 3$ for $x$ — $p(3)$ evaluates the polynomial at $x=3$; solving $p(x)=3$ for $x$ is the reverse question entirely.
If $p(x) = x^2 + 2x - 3$, then $p(-1)$ equals
$0$
$-4$
$-6$
$-1$
Check your answer
$0$ — The middle term is $2 \cdot (-1) = -2$, not $+2$; carrying that sign through gives $1-2-3=-4$, not $0$.
✓ $-4$ — (B) Substituting $x=-1$ gives $1 - 2 - 3 = -4$, once every sign is carried through correctly.
$-6$ — $(-1)^2 = 1$, not $-1$; using the correct square gives $1-2-3=-4$, not $-6$.
$-1$ — $p(-1)$ is the RESULT of evaluating the polynomial, not the input $-1$ itself.
A number $k$ is a zero of $p(x)$ exactly when
$k$ is the value of $p(x)$
$p(x) = k$ for all $x$
$p(k) = 0$
$k$ is positive
Check your answer
$k$ is the value of $p(x)$ — $k$ is the INPUT; whether it is a zero depends on the OUTPUT, $p(k)$, being $0$.
$p(x) = k$ for all $x$ — $p(k)=0$ is a statement about the single point $x=k$, not about $p(x)$ everywhere.
✓ $p(k) = 0$ — (C) The value of the polynomial at $k$ must come out to exactly $0$ for $k$ to be a zero.
$k$ is positive — A zero can be positive, negative, or zero itself; sign has no bearing on whether $p(k)=0$.
Look at $3 x^2 - x + 5$. Find its highest power of $x$. It is $2$. So its DEGREE is $2$.
Let us say the same rule with words. The degree of a polynomial is the highest power of the variable in it.
Your turn: what is the degree of $7 x - 1$? (Answer: $1$, the highest power of $x$.)
Degree gives each polynomial a name. Degree $1$ is LINEAR, written $a x + b$. Degree $2$ is QUADRATIC, written $a x^2 + b x + c$. Degree $3$ is CUBIC. In every one of these, the leading coefficient is not $0$. For the quadratic form, $a \neq 0$ is part of the name, not a separate check we do afterward.
*A quadratic always has $a \neq 0$: drop that, and $a x^2 + b x + c$ is not a quadratic any more, it is the linear polynomial $b x + c$.*
A line crosses the x-axis once, a parabola crosses it twice, and a cubic crosses it three times.
Check yourself
The degree of $p(x) = 5x^3 - 2x^2 + 7$ is
$5$
$3$
$2$
$7$
Check your answer
$5$ — $5$ is the coefficient of $x^3$, not the power itself; the degree is the exponent, $3$.
✓ $3$ — (B) The degree is the highest power of $x$ appearing in the polynomial, which is $3$ here.
$2$ — $2$ is the exponent on the SECOND term; the highest exponent present is $3$, on the first term.
$7$ — $7$ is the constant term, which carries no power of $x$ at all; it cannot be the degree.
A quadratic polynomial is written $a x^2 + b x + c$ with the condition
$b \neq 0$
$c \neq 0$
$a = 1$
$a \neq 0$
Check your answer
$b \neq 0$ — $b$ may be $0$ (as in $x^2-3$); it is $a$ that must stay nonzero to keep the degree at $2$.
$c \neq 0$ — $c$ may be $0$; only $a=0$ would drop the polynomial below degree $2$.
$a = 1$ — A quadratic’s leading coefficient may be any nonzero number, not only $1$.
✓ $a \neq 0$ — (D) If $a$ were $0$ the $x^2$ term would vanish and the polynomial would no longer be degree $2$.
Which of these is a quadratic polynomial?
$3x - 1$
$3x^2 - x + 1$
$3x^3 + x^2 - 1$
$5$
Check your answer
$3x - 1$ — This polynomial’s highest power is $1$, making it linear, not quadratic.
✓ $3x^2 - x + 1$ — (B) Its highest power of $x$ is exactly $2$, which is what makes a polynomial quadratic.
$3x^3 + x^2 - 1$ — This polynomial’s highest power is $3$, making it cubic, not quadratic.
$5$ — A constant carries no power of $x$ at all, so its degree is $0$, not $2$.
Take the line $p(x) = 2 x - 6$. Solve $2 x - 6 = 0$. Add $6$ to both sides: $2 x = 6$. Divide by $2$: $x = 3$. That is the zero.
Now picture the graph of $y = p(x)$, and you find the same zero. At $x = 3$, the curve sits exactly on the x-axis, at the point $(3, 0)$. Every point on the x-axis has $y = 0$. So a zero of $p(x)$ is the x-coordinate of a point where its graph meets the x-axis.
For any linear polynomial $a x + b$ with $a \neq 0$, the graph is a straight, tilted line. A tilted line meets the x-axis at exactly one point. Solve $a x + b = 0$ and we find that point directly: $x = -b/a$.
Your turn: check it on $p(x) = 2 x - 6$. (Answer: $a = 2$, $b = -6$, and $-b/a = 6/2 = 3$, the same zero found above.)
The graph and the algebra find the same zero: read it where the curve meets the x-axis, or solve $p(x) = 0$. Two ways, one answer.
The parabola crosses the x-axis at its two zeros, 1 and 3, exactly where the factored form says.
A SCHOLAR INDIA REMEMBERS
When we write $p(x) = 0$, what is the $0$ telling you? It is the height of the graph. Take $p(x) = x^2 - 4$. At $x = 2$, $p(2) = 4 - 4 = 0$. At $x = -2$, $p(-2) = 4 - 4 = 0$. At both points the graph has height $0$. So both points sit on the x-axis.
Brahmagupta · the brahminy kite
Check yourself
For the linear polynomial $p(x) = 2x - 6$, its zero is the x-coordinate where its graph meets the x-axis, namely
$x = 3$
$x = -3$
$x = 6$
$x = 2$
Check your answer
✓ $x = 3$ — (A) Setting $2x-6=0$ and solving gives $x=3$, the graph’s meeting point with the x-axis.
$x = -3$ — Solving $2x-6=0$ gives $2x=6$, so $x=3$ — not $-3$.
$x = 6$ — $2x=6$ still needs dividing by $2$; the zero is $x=3$, not $6$.
$x = 2$ — $2$ is the coefficient in $2x-6$; it is not the value that makes the expression $0$.
The graph of $y = p(x)$ passes through $(4, 0)$. What does this say about $p(x)$?
$p(x) = 4$ for all $x$
the y-intercept of $p(x)$ is $4$
the degree of $p(x)$ is $4$
$4$ is a zero of $p(x)$
Check your answer
$p(x) = 4$ for all $x$ — One point on a graph does not fix the value of $p(x)$ everywhere; only $p(4)=0$ is known.
the y-intercept of $p(x)$ is $4$ — $(4,0)$ lies on the x-axis; the y-intercept is the separate value $p(0)$.
the degree of $p(x)$ is $4$ — The number $4$ here is a zero’s location, not the highest power of $x$ in $p(x)$.
✓ $4$ is a zero of $p(x)$ — (D) A point with $y=0$ on the graph means the polynomial’s value there is $0$, which is exactly the definition of a zero.
For a linear polynomial $a x + b$ ($a \neq 0$), the graph meets the x-axis
at exactly two points
at $(-b/a, 0)$ only
at the point $(b/a, 0)$
at the origin always
Check your answer
at exactly two points — A straight line meets the x-axis at most once; two meeting points is the quadratic case, not the linear one.
✓ at $(-b/a, 0)$ only — (B) A straight line with a nonzero slope crosses the x-axis exactly once, at $x=-b/a$.
at the point $(b/a, 0)$ — Solving $a x+b=0$ gives $x=-b/a$; dropping the negative sign gives the wrong point.
at the origin always — The meeting point is $(-b/a,0)$ in general; it is the origin only in the special case $b=0$.
A line’s graph meets the x-axis at $(5, 0)$ and nowhere else. If this line is $y = p(x)$ for a linear $p(x) = a x + b$, then $p(x)$ has
two zeros, since a line has two ends
no zero, since the line only touches at one point
exactly one zero, $x = 5$
infinitely many zeros, since the line extends forever
Check your answer
two zeros, since a line has two ends — A line extending in two directions has nothing to do with how many times it crosses the x-axis, which is at most once.
no zero, since the line only touches at one point — Touching the x-axis at one point IS a zero — a linear graph meeting the axis at all gives exactly one zero, not none.
✓ exactly one zero, $x = 5$ — (C) A linear polynomial’s graph meets the x-axis at most once, and here that single meeting point is given directly.
infinitely many zeros, since the line extends forever — How far a line extends says nothing about how many times it meets the x-axis; here it is given as exactly once.
Three points are enough to see the shape every quadratic graph shares: a cup with one turning point and a line of symmetry through it.
Take $y = x^2 - 4$. At $x = 0$, $y = -4$. At $x = 2$, $y = 0$. At $x = -2$, $y = 0$. Plot the three points and join them yourself. You get a smooth, symmetric curve with one turning point, a PARABOLA. Every quadratic $y = a x^2 + b x + c$, with $a \neq 0$, graphs the same way.
Let us say it again with the sign of $a$. When $a > 0$, the parabola opens upward, like a cup, and its turning point is the lowest point on the curve. When $a < 0$, it opens downward, like a dome, and the turning point is the highest point.
Your turn: which way does $y = x^2 - 4$ open? (Answer: upward, because $a = 1 > 0$.)
Where the curve actually crosses the x-axis is a separate question, taken up next.
Two parabolas share the same size of leading coefficient. One opens upward like a cup; the other opens downward like a dome.
Check yourself
The graph of $y = -3x^2 + x - 1$
opens upward, since $-1$ is the constant term
opens downward, since $x$ has a positive coefficient
opens upward, since the highest power is even
opens downward, since $a = -3 < 0$
Check your answer
opens upward, since $-1$ is the constant term — The shape depends on $a$, the coefficient of $x^2$, not on $c$, the constant term.
opens downward, since $x$ has a positive coefficient — The shape depends on $a$, not on $b$, the coefficient of $x$.
opens upward, since the highest power is even — Every quadratic has $x^2$ as its highest power; that alone never decides the direction, only the sign of $a$ does.
✓ opens downward, since $a = -3 < 0$ — (D) The sign of the leading coefficient, $a$, decides whether a parabola opens upward or downward.
A parabola $y = a x^2 + b x + c$ opens upward. What can be said about $a$?
$a > 0$
$a < 0$
$a = 0$
$c > 0$
Check your answer
✓ $a > 0$ — (A) Opening upward is exactly the condition the chapter ties to a positive leading coefficient.
$a < 0$ — $a<0$ is the condition for opening DOWNWARD, the opposite of what is described here.
$a = 0$ — If $a=0$ the polynomial would not be quadratic at all, so it could not form a parabola.
$c > 0$ — $c$ only shifts the graph up or down; it is $a$ that decides whether it opens upward or downward.
Two parabolas are given: $y = 2x^2 - x$ and $y = -5x^2 + 4$. Which one opens downward?
$y = 2x^2 - x$, since it has a negative middle term
$y = -5x^2 + 4$, since its leading coefficient is negative
both open downward, since each has a negative term somewhere
neither opens downward, since both have a squared term
Check your answer
$y = 2x^2 - x$, since it has a negative middle term — The $-x$ term is the middle term, not the leading one; the leading coefficient here is $2$, which is positive.
✓ $y = -5x^2 + 4$, since its leading coefficient is negative — (B) Only the sign of the coefficient of $x^2$ decides the direction, and $-5$ is the one negative leading coefficient here.
both open downward, since each has a negative term somewhere — Having some negative term is not the rule; only the SIGN OF $a$, the $x^2$ coefficient, decides the direction.
neither opens downward, since both have a squared term — A squared term with a negative coefficient still opens downward; the sign of $a$, not the presence of $x^2$, decides this.
Let us look at three graphs.
- Two zeros: $y = x^2 - 4$ crosses the x-axis at $x = -2$ and $x = 2$.
- One zero: $y = (x - 1)^2$ only touches the x-axis, at $x = 1$.
- No zero: $y = x^2 + 1$ never reaches the x-axis at all.
A parabola meets the x-axis in exactly one of these three ways, and no others.
A touch is one repeated zero. It is not zero zeros, because the curve does meet the axis, at $x = 1$. It is not two zeros either, because the curve never crosses to the other side.
Your turn: check it yourself on $y = (x - 1)^2$. Try an $x$ value just below $1$ and one just above $1$. (Answer: both give a positive $y$, so the curve stays on one side of the axis both times.)
One frame with three parabolas sharing the same middle term. One crosses the x-axis twice, one touches it once, one never meets it.
A SCHOLAR INDIA REMEMBERS
Your graph never meets the x-axis. Did you find no zeros, or did you not draw far enough? We do not know yet, so let us check. Take $x^2 + 1$. $x^2$ is never below $0$. So $x^2 + 1$ is at least $1$, for every $x$. It never reaches $0$. There are no zeros to find.
Brahmagupta · the brahminy kite
Check yourself
A parabola can meet the x-axis in
two, one, or zero points
exactly two points, always
one point, always
any number of points
Check your answer
✓ two, one, or zero points — (A) These are the three possible ways a parabola’s curve can relate to the x-axis.
exactly two points, always — A parabola may also touch at one point or miss the axis entirely — two points is only one of three cases.
one point, always — A parabola may also cross at two points or miss the axis entirely — one point is only one of three cases.
any number of points — A quadratic’s graph can meet the x-axis at most twice, never more — “any number” is not possible.
A parabola touches the x-axis at exactly one point without crossing it. This means $p(x)$ has
two distinct zeros
one repeated zero
no real zero, since the curve does not cross the axis
an undefined number of zeros
Check your answer
two distinct zeros — Touching at one point is a single meeting, not two — two distinct zeros would need two separate crossing points.
✓ one repeated zero — (B) A single touch without crossing is exactly the graphical picture of a repeated zero.
no real zero, since the curve does not cross the axis — Touching the axis still counts as meeting it; only a curve that never reaches the axis has no real zero.
an undefined number of zeros — This exact case has a name and a count: one repeated zero, not an undefined outcome.
A student says: “since a parabola is a curve, it must always cross the x-axis twice.” What is wrong with this claim?
nothing is wrong, every parabola crosses twice
a parabola always touches the x-axis exactly once
a parabola only misses the x-axis when $a = 0$
it ignores the miss and the touch cases
Check your answer
nothing is wrong, every parabola crosses twice — Many parabolas touch the axis once or miss it entirely — “always crosses twice” ignores those two real cases.
a parabola always touches the x-axis exactly once — Some parabolas cross at two points and some miss the axis entirely — “always touches once” is just as wrong a restriction.
a parabola only misses the x-axis when $a = 0$ — If $a=0$ the graph would not be a parabola at all; missing the axis can happen for plenty of nonzero $a$.
✓ it ignores the miss and the touch cases — (D) The claim ignores two of the three possible cases a parabola’s graph can show.
A quadratic’s graph never meets the x-axis. What can you conclude about its zeros, considering only real numbers?
it has exactly one zero
it has two zeros, both negative
it has a zero equal to the y-intercept
it has no real zero
Check your answer
it has exactly one zero — One zero corresponds to TOUCHING the axis once; a graph that never meets the axis has no real zero, not one.
it has two zeros, both negative — A graph that never meets the x-axis has no real zeros to be positive or negative — none exist.
it has a zero equal to the y-intercept — The y-intercept is a separate value, $p(0)$; it does not become a zero just because no real zero exists.
✓ it has no real zero — (D) Never meeting the x-axis is exactly the graphical picture of the no-real-zero case.
Take $y = x^2 - 5 x + 6$. Its graph crosses the x-axis at $x = 2$ and $x = 3$. That is two zeros. Its degree is $2$.
Try to add a third crossing to that same graph. We cannot: a quadratic’s degree caps it at $2$ zeros. A cubic, degree $3$, can have at most $3$ zeros. Say it with a letter: a degree-$n$ polynomial has at most $n$ zeros.
This turns counting into something you read straight off a graph, with no algebra at all. Count the points where the curve meets the x-axis. That count is the number of zeros. A single touch point counts once, not twice.
Your turn: a graph crosses the x-axis at four points. What is the smallest degree the polynomial can have? (Answer: $4$, one for each crossing.)
Read the degree as a ceiling on the number of zeros, not a promise that the graph reaches it.
Four small graphs cross the x-axis two, one, zero, and four times, read straight off each picture.
Check yourself
A polynomial of degree $5$ has, at most,
$5$ real zeros
$4$ real zeros
$10$ real zeros
an unlimited number of real zeros
Check your answer
✓ $5$ real zeros — (A) The degree itself is the ceiling on how many zeros a polynomial’s graph can show.
$4$ real zeros — The bound equals the degree exactly, $5$, not one less than it.
$10$ real zeros — The maximum number of zeros equals the degree itself, $5$, not double it.
an unlimited number of real zeros — The degree fixes a definite ceiling; a degree-$5$ polynomial cannot have more than $5$ real zeros.
The graph of $y = p(x)$ meets the x-axis at $4$ distinct points. What is the smallest possible degree of $p(x)$?
$2$
$4$
$8$
$3$
Check your answer
$2$ — A quadratic can show at most $2$ zeros; $4$ distinct crossings need a degree of at least $4$.
✓ $4$ — (B) Each distinct crossing point is a distinct zero, so the degree must be at least as large as the count shown.
$8$ — The degree needed is at least the number of zeros shown, $4$, not double that count.
$3$ — Four distinct zeros require a degree of at least $4$; a degree-$3$ polynomial could show at most $3$.
Why can counting the crossing points on a graph tell you the number of zeros, without any algebra?
because every polynomial’s graph is a straight line
because the degree always equals the number of crossings, whatever the graph shows
each crossing point is, by definition, a value where the polynomial equals zero
because zeros can only be found by counting graph crossings, never by algebra
Check your answer
because every polynomial’s graph is a straight line — Only degree-$1$ polynomials graph as straight lines; the counting method works for graphs of any shape.
because the degree always equals the number of crossings, whatever the graph shows — The degree only sets a MAXIMUM; the actual number of crossings can be fewer, so the two are not always equal.
✓ each crossing point is, by definition, a value where the polynomial equals zero — (C) A crossing point is exactly where the function value is $0$, which is the definition of a zero, so counting crossings counts zeros directly.
because zeros can only be found by counting graph crossings, never by algebra — Splitting the middle term also finds zeros, algebraically; counting crossings is a second, graphical way, not the only way.
A rectangle of area x squared plus 2x plus 3x plus 6 splits into sides x plus 2 and x plus 3.
Take $x^2 + 5 x + 6$. We need two numbers that multiply to $6$ and add to $5$. Those numbers are $2$ and $3$. This is splitting the middle term. It turns a quadratic into two factors, and each factor hands over one zero for free.
Say it again for any quadratic. For $a x^2 + b x + c$, find two numbers that multiply to $a c$ and add to $b$. Use those two numbers to split the middle term $b x$ into two pieces. We now have four terms instead of three.
Group the four terms in pairs, and pull the common factor out of each pair. The same bracket should appear behind both pairs. That shared bracket is one factor, and what is left outside it is the other.
Your turn: split $x^2 + 5 x + 6$ this way. Group $x^2 + 2 x + 3 x + 6$ into two pairs. (Answer: $x(x + 2) + 3(x + 2)$, and $(x + 2)$ appears both times.)
*Each zero of the quadratic is the value of $x$ that makes one of the two factors equal to $0$.* Find the right pair of numbers first. Guessing at the grouping stage wastes the work already done.
Check yourself
To split the middle term of $a x^2 + b x + c$, the two new coefficients must
multiply to $b$ and add to $a c$
multiply to $a c$ and add to $b$
add to $a c$ and multiply to $b$
multiply to $c$ and add to $a$
Check your answer
multiply to $b$ and add to $a c$ — It is the PRODUCT that must equal $a c$ and the SUM that must equal $b$ — this swaps the two.
✓ multiply to $a c$ and add to $b$ — (B) These two conditions together are exactly what the splitting method requires.
add to $a c$ and multiply to $b$ — The sum must equal $b$ and the product must equal $a c$ — this reverses both roles.
multiply to $c$ and add to $a$ — The correct target product is $a c$, using BOTH $a$ and $c$ together, not $c$ alone paired with $a$.
To split the middle term of $x^2 + 7x + 10$, the two numbers needed are
$1$ and $10$
$2$ and $5$
$-2$ and $-5$
$7$ and $10$
Check your answer
$1$ and $10$ — $1 \cdot 10 = 10$ is right, but $1+10=11$, not the required sum of $7$.
✓ $2$ and $5$ — (B) $2$ and $5$ multiply to $10$ and add to $7$, exactly the two conditions the method sets.
$-2$ and $-5$ — $(-2) \cdot (-5)=10$ is right, but $(-2)+(-5)=-7$, not $7$.
$7$ and $10$ — $7$ and $10$ are the coefficients already given; the method needs a NEW pair that multiplies to $10$ and adds to $7$.
Splitting $a x^2 + b x + c$’s middle term into four terms sets up factorisation because
the four terms can then be added directly to get the answer
the four terms must then be multiplied together
each grouped pair shares a common factor
two of the four terms are then dropped as unnecessary
Check your answer
the four terms can then be added directly to get the answer — The four terms are grouped in PAIRS to expose a shared factor, not simply added together.
the four terms must then be multiplied together — Multiplying all four terms together has no role in this method; the terms are grouped in pairs, not multiplied.
✓ each grouped pair shares a common factor — (C) The whole point of the split is that each resulting pair of terms shares a factor, which is what lets the expression be grouped into two binomials.
two of the four terms are then dropped as unnecessary — Every one of the four terms is kept and used in a pair; none are dropped.
Using the same splitting idea as $2x^2 - 8x + 6 = 2(x-1)(x-3)$ (worked in this chapter), what two numbers split the middle term of $3x^2 - 11x + 6$?
$-3$ and $-6$
$2$ and $9$
$-2$ and $-9$
$-1$ and $-18$
Check your answer
$-3$ and $-6$ — $(-3) \cdot (-6)=18$ matches $a c$, but $(-3)+(-6)=-9$, not the required $-11$.
$2$ and $9$ — $2 \cdot 9=18$ matches $a c$, but $2+9=11$, the wrong sign for the required $-11$.
✓ $-2$ and $-9$ — (C) $-2$ and $-9$ multiply to $18$, which is $a c$ for this quadratic, and add to $-11$, which is $b$.
$-1$ and $-18$ — $(-1) \cdot (-18)=18$ matches $a c$, but $(-1)+(-18)=-19$, not $-11$.
Two parabolas share the same axis of symmetry, so both pairs of zeros add to the same sum, 2, even though their products differ.
Take $x^2 - 5 x + 6$, with zeros $2$ and $3$. Add them: $2 + 3 = 5$. Multiply them: $2 \cdot 3 = 6$.
Now check both against the coefficients. $-b/a = -(-5)/1 = 5$, and that matches the sum. $c/a = 6/1 = 6$, and that matches the product. Once we know a quadratic’s coefficients, we can read off its zeros’ sum and product directly. No factorising is needed first.
Say it again for any quadratic. For $a x^2 + b x + c$ with $a \neq 0$ and zeros $\alpha$, $\beta$: the sum $\alpha + \beta$ equals $-b/a$, and the product $\alpha \beta$ equals $c/a$.
Your turn: for $x^2 + x - 12 = (x + 4)(x - 3)$, the zeros are $-4$ and $3$. (Answer: sum $-4 + 3 = -1$, matching $-b/a = -1/1 = -1$; product $-4 \cdot 3 = -12$, matching $c/a = -12/1 = -12$.)
The minus sign belongs to the sum only, never to the product.
Check yourself
For $a x^2 + b x + c$ with zeros $\alpha, \beta$, the sum $\alpha + \beta$ equals
$-b/a$
$b/a$
$-c/a$
$a/b$
Check your answer
✓ $-b/a$ — (A) The sum of the zeros reads off the coefficients as minus the coefficient of $x$ over the coefficient of $x^2$.
$b/a$ — The sum is $-b/a$, not $b/a$ — the negative sign is part of the formula, not optional.
$-c/a$ — $-c/a$ uses the wrong coefficient; the sum uses $b$, the coefficient of $x$, not $c$.
$a/b$ — The formula is $-b/a$, with $b$ over $a$; inverting it to $a/b$ is not the same value in general.
For $a x^2 + b x + c$ with zeros $\alpha, \beta$, the product $\alpha \beta$ equals
$-c/a$
$c/a$
$b/a$
$a/c$
Check your answer
$-c/a$ — The product is $c/a$, with no negative sign — that belongs to the sum formula, not this one.
✓ $c/a$ — (B) The product of the zeros reads off the coefficients as the constant term over the coefficient of $x^2$.
$b/a$ — $b/a$ uses the wrong coefficient; the product uses $c$, the constant term, not $b$.
$a/c$ — The formula is $c/a$, with $c$ over $a$; inverting it to $a/c$ is not the same value in general.
For $p(x) = 2x^2 - 7x + 3$, the sum of the zeros is
$7/2$
$-7/2$
$3/2$
$2/7$
Check your answer
✓ $7/2$ — (A) $-b/a = -(-7)/2 = 7/2$, once the double negative from $b=-7$ is carried through correctly.
$-7/2$ — $b=-7$, so $-b/a = -(-7)/2 = 7/2$; forgetting that $b$ is already negative leaves the wrong sign.
$3/2$ — $3/2$ is $c/a$, the PRODUCT of the zeros; the sum uses $-b/a$ instead.
$2/7$ — The sum is $-b/a=7/2$; inverting the coefficients to $2/7$ does not compute the formula correctly.
Why does the sum of a quadratic’s zeros equal $-b/a$ rather than simply $-b$?
because $b$ is already the sum before any factor of $a$ is considered
because zeros are always fractions, never whole numbers
because $a$ has no effect on the sum or product of the zeros
because $a$ must be divided out to isolate the sum
Check your answer
because $b$ is already the sum before any factor of $a$ is considered — $b$ equals $-a(\alpha+\beta)$, not $\alpha+\beta$ directly; dividing by $a$ is required to isolate the sum.
because zeros are always fractions, never whole numbers — Zeros can be whole numbers, as in earlier examples in this chapter; this has nothing to do with why $a$ appears in the formula.
because $a$ has no effect on the sum or product of the zeros — $a$ appears in the denominator of both $-b/a$ and $c/a$ — it directly affects both the sum and the product.
✓ because $a$ must be divided out to isolate the sum — (D) The leading coefficient $a$ scales the whole factored form, so it must be divided out before $b$ alone matches the sum.
A student computes the zeros’ sum and product for $p(x) = x^2 + 5x - 6$ as: sum $= 5$, product $= -6$. What went wrong?
the product should be $6$, not $-6$
both values are correct as computed
the sum should be $-6$ and the product $5$, the two values were swapped
the sum should be $-5$, not $5$ — the sign of $b$ was not flipped
Check your answer
the product should be $6$, not $-6$ — The product $c/a=-6/1=-6$ is already correct — the error is in the sum, not the product.
both values are correct as computed — The sum should be $-b/a=-5$; the student’s $5$ drops the required negative sign.
the sum should be $-6$ and the product $5$, the two values were swapped — Swapping the two values does not fix the real issue: the sum formula needs its negative sign restored, giving $-5$, not $-6$.
✓ the sum should be $-5$, not $5$ — the sign of $b$ was not flipped — (D) The sum formula is $-b/a$, and here $b=5,a=1$, so the sum is $-5$; the student used $b$ directly instead of $-b$.
Suppose we want a quadratic whose zeros add to $-3$ and multiply to $2$. Try $x^2 - (-3) x + 2 = x^2 + 3 x + 2$. Factorise it: $(x + 1)(x + 2)$, with zeros $-1$ and $-2$. You can check: they add to $-3$ and multiply to $2$, exactly as wanted.
Say it again with letters. Given any target sum $s$ and product $p$, the polynomial $x^2 - s x + p$ has exactly those zeros.
Multiply that polynomial by any nonzero constant $k$. The result, $k x^2 - k s x + k p$, has the same pair of zeros. The constant scales every term equally, so it cancels out of both the sum and the product.
Your turn: build a quadratic whose zeros add to $5$ and multiply to $6$. (Answer: $x^2 - 5 x + 6$.)
Three parabolas, scaled by 0.5, 1 and 2, all cross the x-axis at the same two zeros, 1 and 4.
Check yourself
Given a target sum $s$ and product $p$ for the zeros, a matching quadratic is
$x^2 + s x + p$
$x^2 - s x + p$
$x^2 - s x - p$
$x^2 - p x + s$
Check your answer
$x^2 + s x + p$ — The construction needs $-s x$, not $+s x$ — the sign matches the sum formula $-b/a$ run in reverse.
✓ $x^2 - s x + p$ — (B) This is exactly the construction that runs the sum-product relation in reverse.
$x^2 - s x - p$ — The constant term should be $+p$, not $-p$, to match the product formula $c/a$ run in reverse.
$x^2 - p x + s$ — The coefficient of $x$ should use the SUM, $s$, and the constant should use the PRODUCT, $p$ — this swaps the two.
A quadratic whose zeros have sum $4$ and product $-3$ is
$x^2 + 4x - 3$
$x^2 - 4x + 3$
$x^2 - 4x - 3$
$x^2 - 3x + 4$
Check your answer
$x^2 + 4x - 3$ — With $s=4$, the middle term should be $-4x$, not $+4x$.
$x^2 - 4x + 3$ — With $p=-3$, the constant term should stay $-3$, not become $+3$.
✓ $x^2 - 4x - 3$ — (C) Using $x^2-s x+p$ with $s=4,p=-3$ gives exactly $x^2-4x-3$.
$x^2 - 3x + 4$ — The sum, $4$, belongs in the $x$-coefficient position, and the product, $-3$, in the constant position — this swaps them.
If $x^2 - s x + p$ has zeros with sum $s$ and product $p$, what happens to the zeros of $3(x^2 - s x + p) = 3x^2 - 3 s x + 3p$?
the zeros are multiplied by $3$
the zeros stay exactly the same
the sum becomes $3s$ but the product stays $p$
the new quadratic has no real zeros
Check your answer
the zeros are multiplied by $3$ — Multiplying the whole polynomial by $3$ scales its VALUES, not the $x$ values where it equals $0$; the zeros stay the same.
✓ the zeros stay exactly the same — (B) Scaling every term by a nonzero constant does not change which $x$ values make the expression $0$, so the zeros are unaffected.
the sum becomes $3s$ but the product stays $p$ — Since the zeros themselves do not change, both their sum and their product stay exactly as they were, $s$ and $p$.
the new quadratic has no real zeros — Scaling by a nonzero constant preserves every zero the original quadratic had — it cannot remove them.
Two quadratics, $x^2 - 5x + 6$ and $2x^2 - 10x + 12$, are given. What is true about their zeros?
the second quadratic’s zeros are double the first’s
the second quadratic has no zeros in common with the first
only the first quadratic’s zeros are real
both have the same pair of zeros, $2$ and $3$
Check your answer
the second quadratic’s zeros are double the first’s — Scaling the whole polynomial by $2$ does not double where it equals $0$; the zeros themselves stay $2$ and $3$ in both cases.
the second quadratic has no zeros in common with the first — The second quadratic is a constant multiple of the first, so by the scaling rule they share exactly the same zeros.
only the first quadratic’s zeros are real — A nonzero constant multiple of a quadratic has exactly the same zeros, real or not, as the original.
✓ both have the same pair of zeros, $2$ and $3$ — (D) The second quadratic is exactly $2$ times the first, and scaling by a nonzero constant never changes the zeros.
Take $p(x) = x^2 - 3 x - 4$. Its zeros are $-1$ and $4$: $p(-1) = 0$ and $p(4) = 0$. Both points sit on the x-axis.
Now we look at $x = 0$. $p(0) = -4$. That value, $-4$, is where the curve crosses the y-axis, the Y-INTERCEPT of $p(x)$.
A zero and a y-intercept are two different questions about the same graph. A zero asks where the curve meets the x-axis. The y-intercept asks only what $p(0)$ is. It is easy to reach for the wrong one, because both are single numbers read straight off a graph.
Your turn: is $-4$, the y-intercept above, also a zero of $p(x)$? (Answer: no. $p(-4) = 16 + 12 - 4 = 24$, not $0$.)
The graph shows three points: the two zeros, at minus 1 and 4, and the y-intercept, at minus 4.
Is the constant term always a zero you can read straight off a quadratic?
Weaker. A student is asked for a zero of $p(x) = x^2 - x - 6$. They look at the constant term. “The zero is $-6$,” they say. “That is where the graph crosses the axis.”
They do not check it. Put $-6$ into $p(x)$: $p(-6) = 36 + 6 - 6 = 36$, not $0$. $-6$ is not a zero at all.
Stronger. Factorise first. $x^2 - x - 6 = (x - 3)(x + 2)$, so the zeros are $3$ and $-2$: $p(3) = 0$ and $p(-2) = 0$.
The constant term, $-6$, is the y-intercept, $p(0) = -6$. It answers a different question: where the curve crosses the y-axis, not the x-axis.
A SCHOLAR INDIA REMEMBERS
Two different zeros are at work here. Keep them apart. Take $p(x) = x^2 - 5x + 6$. Put $x = 0$ and you get $6$. That is where the graph crosses the y-axis. Now put $p(x) = 0$. You get $x = 2$ and $x = 3$. Those are the zeros. One zero goes in, the other comes out.
Brahmagupta · the brahminy kite
Check yourself
A student claims: “the zero of a polynomial $p(x)$ is the point where its graph crosses the y-axis.” Is this correct?
yes, since both describe a point where the graph meets an axis
no, because a zero is always the same number as the y-intercept
no — it crosses the x-axis, not the y-axis
no, because a polynomial’s graph never touches the y-axis
Check your answer
yes, since both describe a point where the graph meets an axis — Meeting SOME axis is not the point — a zero specifically meets the x-axis, and the y-axis crossing is a different value, $p(0)$.
no, because a zero is always the same number as the y-intercept — A zero and the y-intercept are usually different numbers; they coincide only in the special case where $0$ itself happens to be a zero.
✓ no — it crosses the x-axis, not the y-axis — (C) A zero is defined by $p(k)=0$, which is a crossing of the x-axis; the y-axis crossing is the separate y-intercept.
no, because a polynomial’s graph never touches the y-axis — Every polynomial’s graph crosses the y-axis exactly once, at $p(0)$; that crossing is simply not the same thing as a zero.
For $p(x) = x^2 - 3x - 4$, the zeros are $-1$ and $4$. What is the y-intercept?
$p(0) = -4$
$-1$, since that is one of the zeros
$4$, since that is the other zero
$0$, since the y-intercept is where $x = 0$ crosses the curve
Check your answer
✓ $p(0) = -4$ — (A) The y-intercept is found by substituting $x=0$, which gives $-4$ here, a value separate from either zero.
$-1$, since that is one of the zeros — $-1$ is a zero, found where $p(x)=0$; the y-intercept is the different value $p(0)=-4$.
$4$, since that is the other zero — $4$ is the other zero; the y-intercept is still the separate value $p(0)=-4$.
$0$, since the y-intercept is where $x = 0$ crosses the curve — $x=0$ is the INPUT that gives the y-intercept; the y-intercept itself is the OUTPUT, $p(0)=-4$.
A student checks whether $-4$ (the y-intercept of $p(x)=x^2-3x-4$) is also a zero, by computing $p(-4) = 16+12-4=24$. What does this show?
$-4$ must be a zero, since it is the y-intercept
$-4$ is not a zero, since $p(-4) \neq 0$
the computation is wrong, since the y-intercept is always a zero
$p(x)$ has no zeros at all, since $-4$ failed the check
Check your answer
$-4$ must be a zero, since it is the y-intercept — The very computation shown, $p(-4)=24 \neq 0$, proves $-4$ is NOT a zero, whatever it is as the y-intercept.
✓ $-4$ is not a zero, since $p(-4) \neq 0$ — (B) The computation gives a nonzero result, which directly shows $-4$ fails the definition of a zero despite being the y-intercept.
the computation is wrong, since the y-intercept is always a zero — The computation $p(-4)=24$ is correct; the false idea is the rule that a y-intercept must be a zero.
$p(x)$ has no zeros at all, since $-4$ failed the check — This chapter already found $p(x)$’s real zeros, $-1$ and $4$; one non-zero value, $-4$, says nothing about those.
For which of these is the y-intercept also a zero?
$p(x) = x^2 - 3x - 4$, since it has a y-intercept
$p(x) = x^2 - 5x$, since $p(0) = 0$
$p(x) = x^2 + 1$, since it is a simple quadratic
no polynomial’s y-intercept can ever be a zero
Check your answer
$p(x) = x^2 - 3x - 4$, since it has a y-intercept — Every polynomial has SOME y-intercept; that alone never makes it a zero — here $p(0)=-4 \neq 0$.
✓ $p(x) = x^2 - 5x$, since $p(0) = 0$ — (B) Here the y-intercept and a zero happen to coincide, because substituting $x=0$ gives exactly $0$.
$p(x) = x^2 + 1$, since it is a simple quadratic — Being “simple” has no bearing on this; in fact $p(0)=1 \neq 0$ here, so the y-intercept is not a zero.
no polynomial’s y-intercept can ever be a zero — The y-intercept is a zero exactly when $p(0)=0$ — a real possibility, as $p(x)=x^2-5x$ shows.
Take $p(x) = 2 x^2 - 8 x + 6$, with zeros $1$ and $3$. Add them: $1 + 3 = 4$. That uses $a$: $-b/a = -(-8)/2 = 4$. Multiply them: $1 \cdot 3 = 3$. That uses $a$ too: $c/a = 6/2 = 3$.
We now turn to $b$ and $c$ alone, without $a$. $-b = 8$ and $c = 6$. Those numbers are not the sum and the product here. Dividing by $a$ is not optional. It is part of both relations.
Your turn: for $p(x) = 3 x^2 - 9 x + 6$, what is $a$? (Answer: $a = 3$, so both the sum and the product must be divided by $3$.)
*Sum $= -b/a$. Product $= c/a$. Never $-b$ and $c$ on their own, unless $a = 1$.*
First try
For $2 x^2 - 8 x + 6$ I took the sum of the zeros as $8$ and the product as $6$.
Second look
Those numbers are $-b$ and $c$ only when $a = 1$. Here $a = 2$, so the sum is $-(-8)/2 = 4$ and the product is $6/2 = 3$. Both match the zeros $1$ and $3$.
The sum of the zeros is $-b/a$ and the product is $c/a$. Never drop the leading coefficient $a$.
Do the sum and product ever skip dividing by the leading coefficient?
Weaker. A student is asked for the sum and product of the zeros of $p(x) = 3 x^2 - 9 x + 6$. They write sum $= -b = 9$ and product $= c = 6$. “That is what the rule says,” they say, pointing at $-b$ and $c$ straight from the coefficients.
This feels right, because it worked on every earlier example, where $a$ happened to be $1$. It is wrong here, because $a = 3$, not $1$.
Stronger. Look at $a$ first. Here $a = 3$. Sum $= -b/a = -(-9)/3 = 3$. Product $= c/a = 6/3 = 2$.
Check it against the zeros themselves: $3 x^2 - 9 x + 6 = 3 (x - 1)(x - 2)$, whose zeros are $1$ and $2$. They add to $3$ and multiply to $2$, exactly as the coefficients say.
Zeros of $2 x^2 - 8 x + 6$ by splitting the middle term
$(-6) + (-2) = -8$ and $(-6) \cdot (-2) = 12 = 2 \cdot 6$ We need two numbers that multiply to $a c = 2 \cdot 6 = 12$ and add to $b = -8$. Those numbers are $-6$ and $-2$.
$2 x^2 - 8 x + 6 = 2 x^2 - 6 x - 2 x + 6$ Split the middle term into the two pieces we just found.
$= 2 x (x - 3) - 2 (x - 3)$ Group the four terms in pairs. Pull the common factor out of each pair.
$= (2 x - 2)(x - 3) = 2 (x - 1)(x - 3)$ The shared bracket $(x - 3)$ factors out. What is left behind is the second factor.
$x = 1$ or $x = 3$ Each zero makes one factor equal to $0$.
$p(1) = 2 - 8 + 6 = 0$, $p(3) = 18 - 24 + 6 = 0$ Check both zeros yourself. Put them back into the original polynomial. Both give $0$, so $1$ and $3$ really are the zeros.
Find the zeros of any quadratic by splitting the middle term.
Find the two numbers For $2 x^2 - 8 x + 6$, find two numbers that multiply to $a c = 2 \cdot 6 = 12$ and add to $b = -8$. Those numbers are $-6$ and $-2$.
Split the middle term Rewrite $-8 x$ as $-6 x - 2 x$. $2 x^2 - 8 x + 6 = 2 x^2 - 6 x - 2 x + 6$.
Group in pairs Group the first two terms and the last two. Pull the common factor out of each pair. $2 x (x - 3) - 2 (x - 3)$.
Factor out the shared bracket The bracket $(x - 3)$ appears in both pairs. $(2 x - 2)(x - 3) = 2 (x - 1)(x - 3)$.
Read off the zeros Each factor equal to $0$ gives a zero. $x = 1$ or $x = 3$. Check: $p(1) = 2 - 8 + 6 = 0$, $p(3) = 18 - 24 + 6 = 0$.
rising from the kick
height zero, the ground
it lands here
The ball rises from Meera’s kick and comes back down to the ground, where its height is zero.
Check yourself
In factorising $2x^2 - 8x + 6$, the middle term $-8x$ is split into
$-3x$ and $-5x$
$-6x$ and $-2x$
$-1x$ and $-7x$
$-4x$ and $-4x$
Check your answer
$-3x$ and $-5x$ — $-3+(-5)=-8$ matches the sum, but $(-3) \cdot (-5)=15$, not the required product $12$.
✓ $-6x$ and $-2x$ — (B) $-6$ and $-2$ multiply to $12$, which is $a c=2 \cdot 6$, and add to $-8$, matching $b$.
$-1x$ and $-7x$ — $-1+(-7)=-8$ matches the sum, but $(-1) \cdot (-7)=7$, not the required product $12$.
$-4x$ and $-4x$ — $-4+(-4)=-8$ matches the sum, but $(-4) \cdot (-4)=16$, not the required product $12$.
After splitting, $2x^2-8x+6$ becomes $2x^2-6x-2x+6$, which groups into
$2x(x-3) - 2(x-3)$
$2x(x-3) + 2(x-3)$
$2x(x+3) - 2(x+3)$
$(2x-6)(x-2)$
Check your answer
✓ $2x(x-3) - 2(x-3)$ — (A) Grouping the first pair and the second pair each pulls out a common factor, leaving $(x-3)$ shared by both.
$2x(x-3) + 2(x-3)$ — The second pair is $-2x+6=-2(x-3)$, with a minus sign, not a plus.
$2x(x+3) - 2(x+3)$ — Factoring $2x-6$ gives $2(x-3)$, with a minus inside the bracket, not $(x+3)$.
$(2x-6)(x-2)$ — This skips the grouping step entirely and lands on a factorisation that does not even expand back to $2x^2-8x+6$.
Using the same method as $2x^2-8x+6=2(x-1)(x-3)$, factorise $3x^2-9x+6=3(x^2-3x+2)$. The zeros of $x^2-3x+2$ are
$1$ and $3$
$1$ and $2$
$-1$ and $-2$
$3$ and $6$
Check your answer
$1$ and $3$ — $1$ and $3$ were the zeros of the earlier example, $2x^2-8x+6$; this is a different quadratic with zeros $1$ and $2$.
✓ $1$ and $2$ — (B) $x^2-3x+2$ splits as $(x-1)(x-2)$, giving zeros $1$ and $2$, by the same splitting method used in the worked example.
$-1$ and $-2$ — Factorising gives $(x-1)(x-2)$, so the zeros are $+1$ and $+2$, not their negatives.
$3$ and $6$ — $3$ and $6$ are coefficients in $3x^2-9x+6$, not the zeros — the zeros come from factorising, and are $1$ and $2$.
The coefficients leave a mark on the picture: minus b over a fixes where the two zeros balance, and the curve folds over that line.
Worked example
Where the sum-product relation comes from
$a x^2 + b x + c = k (x - \alpha)(x - \beta)$ Any quadratic with zeros $\alpha$, $\beta$ is some nonzero constant $k$ times these two factors.
$k (x - \alpha)(x - \beta) = k x^2 - k (\alpha + \beta) x + k \alpha \beta$ Expand the right-hand side.
$a = k$ Compare the coefficients of $x^2$ on both sides.
$b = -k (\alpha + \beta)$ Compare the coefficients of $x$.
$c = k \alpha \beta$ Compare the constant terms.
$\alpha + \beta = -b/a$, $\alpha \beta = c/a$ *Put $k = a$ into the last two equations, and solve for the sum and product yourself.* This is where $-b/a$ and $c/a$ come from.
$1 + 3 = 4 = -(-8)/2$ and $1 \cdot 3 = 3 = 6/2$ *Check the new relation against the zeros already found for $2 x^2 - 8 x + 6$.* Sum $4$ and product $3$ match $-b/a$ and $c/a$ exactly.
Check yourself
In deriving the sum-product relation, $a x^2+b x+c$ is first written as
$k(x+\alpha)(x+\beta)$
$k(x-\alpha)(x-\beta)$
$(x-\alpha)(x-\beta)$ with no constant $k$
$k(x-\alpha) + (x-\beta)$
Check your answer
$k(x+\alpha)(x+\beta)$ — A zero at $\alpha$ needs the factor $(x-\alpha)$, not $(x+\alpha)$, so that substituting $x=\alpha$ gives $0$.
✓ $k(x-\alpha)(x-\beta)$ — (B) Writing the quadratic in this factored form, scaled by a constant $k$, is the starting point the whole derivation expands from.
$(x-\alpha)(x-\beta)$ with no constant $k$ — Without $k$, this form can only ever equal $x^2-(\alpha+\beta)x+\alpha \beta$, which forces $a=1$ — the general case needs $k$.
$k(x-\alpha) + (x-\beta)$ — A quadratic’s two zeros come from a PRODUCT of two linear factors, not a sum of two separate expressions.
Expanding $k(x-\alpha)(x-\beta)$ gives $k x^2 - k(\alpha+\beta)x + k \alpha \beta$. Comparing this to $a x^2+b x+c$ term by term gives
$a = k$, $b = k(\alpha+\beta)$, $c = k \alpha \beta$ — The expanded form has $-k(\alpha+\beta)x$, with a minus sign, so $b$ must carry that same minus sign.
$a = -k$, $b = k(\alpha+\beta)$, $c = -k \alpha \beta$ — The $x^2$ coefficient in the expansion is $+k$, not $-k$; flipping every sign does not match the expanded terms.
$a = k \alpha \beta$, $b = -k(\alpha+\beta)$, $c = k$ — $a$ matches the $x^2$ term’s coefficient, which is $k$, not $k \alpha \beta$ — that value belongs to $c$ instead.
✓ $a = k$, $b = -k(\alpha+\beta)$, $c = k \alpha \beta$ — (D) Matching each power of $x$ in turn reads the coefficients straight off the expanded form.
The derivation shows $b=-k(\alpha+\beta)$ and $a=k$. Dividing the first by the second gives $b/a = -(\alpha+\beta)$, which rearranges to
$\alpha + \beta = -b/a$
$\alpha + \beta = b/a$
$\alpha + \beta = -a/b$
$\alpha - \beta = -b/a$
Check your answer
✓ $\alpha + \beta = -b/a$ — (A) Multiplying both sides of $b/a=-(\alpha+\beta)$ by $-1$ gives exactly the sum formula stated at the start of the chapter.
$\alpha + \beta = b/a$ — Rearranging $b/a=-(\alpha+\beta)$ needs multiplying by $-1$, which keeps the negative sign in the final formula.
$\alpha + \beta = -a/b$ — The rearrangement only flips the SIGN of $b/a$; it does not invert the ratio to $a/b$.
$\alpha - \beta = -b/a$ — The derivation is about $\alpha+\beta$ throughout; nothing in it produces a difference, $\alpha-\beta$.
$x^2 - 3 = (x - \sqrt{3})(x + \sqrt{3})$ $x^2 - 3$ is a difference of squares, $x^2 - (\sqrt{3})^2$. Factor it using $a^2 - b^2 = (a - b)(a + b)$.
$x = \sqrt{3}$ or $x = -\sqrt{3}$ Each zero makes one factor equal to $0$.
sum $= \sqrt{3} + (-\sqrt{3}) = 0$ Add the two zeros.
$-b/a = -0/1 = 0$ *This polynomial has no $x$ term, so $b = 0$.* The sum matches.
product $= \sqrt{3} \cdot (-\sqrt{3}) = -3$ Multiply the two zeros.
$c/a = -3/1 = -3$ The product matches too. Check it yourself. Both zeros are irrational, but the sum and the product still come out as whole numbers.
A number line marks the two zeros of the quadratic, the square root of 3 and the same distance on the other side of 0.
Find and check the zeros of a quadratic with no middle term.
Write it as a difference of squares $x^2 - 3$ is $x^2 - (\sqrt{3})^2$. Factor it as $(x - \sqrt{3})(x + \sqrt{3})$.
Read off the zeros Each factor equal to $0$ gives a zero. $x = \sqrt{3}$ or $x = -\sqrt{3}$.
Add the zeros and check $\sqrt{3} + (-\sqrt{3}) = 0$. Since $b = 0$ here, $-b/a = -0/1 = 0$. It matches.
Multiply the zeros and check $\sqrt{3} \cdot (-\sqrt{3}) = -3$. And $c/a = -3/1 = -3$. It matches too.
Check yourself
$p(x) = x^2 - 3$ factorises, using $a^2-b^2=(a-b)(a+b)$, as
$(x-3)(x+1)$
$(x-\sqrt{3})(x-\sqrt{3})$
$(x-\sqrt{3})(x+\sqrt{3})$
$(x-9)(x+9)$
Check your answer
$(x-3)(x+1)$ — This pair does not even expand back to $x^2-3$; the pattern here needs $\sqrt{3}$, not $3$ and $1$.
$(x-\sqrt{3})(x-\sqrt{3})$ — The difference-of-squares pattern needs one $+$ and one $-$ factor; repeating $(x-\sqrt{3})$ twice does not expand to $x^2-3$.
✓ $(x-\sqrt{3})(x+\sqrt{3})$ — (C) Treating $3$ as $(\sqrt{3})^2$ lets the difference-of-squares pattern apply directly.
$(x-9)(x+9)$ — The pattern uses $\sqrt{3}$, the SQUARE ROOT of $3$, not $9$, which is $3$ squared.
For $p(x) = x^2 - 3$, there is no $x$ term, so in $a x^2+b x+c$ this means
$b = 0$
$a = 0$
$c = 0$
$b = 3$
Check your answer
✓ $b = 0$ — (A) A missing $x$ term means its coefficient is simply $0$.
$a = 0$ — If $a=0$ there would be no $x^2$ term at all, and $p(x)$ would not be quadratic.
$c = 0$ — The constant term here is $-3$, which is present and nonzero; it is the $x$ term that is missing.
$b = 3$ — The constant term is $-3$; the coefficient of $x$, $b$, is separately $0$ since no $x$ term appears.
For $p(x) = x^2 - 3$, the sum-product relation predicts sum $=-b/a=0$ and product $=c/a=-3$. Do the zeros $\sqrt{3}$ and $-\sqrt{3}$ match this?
no — the sum should be $\sqrt{3}$, not $0$
no — the product should be $3$, not $-3$
yes — both the sum and product check out
yes, but only because both zeros happen to be irrational
Check your answer
no — the sum should be $\sqrt{3}$, not $0$ — $\sqrt{3}+(-\sqrt{3})=0$ exactly, once the second zero’s negative sign is included; it is not $\sqrt{3}$ alone.
no — the product should be $3$, not $-3$ — $\sqrt{3} \cdot (-\sqrt{3})=-3$; multiplying a positive number by a negative one keeps the result negative.
✓ yes — both the sum and product check out — (C) Adding and multiplying the two zeros directly gives exactly the values the relation predicted.
yes, but only because both zeros happen to be irrational — The relation holds for every quadratic’s zeros, rational or irrational — irrationality plays no special role here.
$x^2 + 3x - 2$ — The constant term should stay $+p=+2$; flipping it to $-2$ does not match the given product.
$x^2 - 3x - 2$ — Only the middle term’s sign flips because $s$ is negative; the constant term should stay $+2$.
Checking $x^2+3x+2=(x+1)(x+2)$ by factorising, the zeros are
$1$ and $2$
$-1$ and $2$
$3$ and $2$
$-1$, $-2$
Check your answer
$1$ and $2$ — $(x+1)=0$ gives $x=-1$, not $x=1$ — the zero is the value that makes the FACTOR zero, with the sign flipped.
$-1$ and $2$ — Both factors need the same sign-flip treatment: $(x+1)=0$ gives $x=-1$ and $(x+2)=0$ gives $x=-2$.
$3$ and $2$ — $3$ and $2$ are coefficients of $x^2+3x+2$, not its zeros — the zeros come from the factors, $-1$ and $-2$.
✓ $-1$, $-2$ — (D) Setting each factor to $0$ gives $x=-1$ and $x=-2$, the actual zeros.
For the zeros $-1, -2$ found above, does their sum and product match the original target of sum $-3$, product $2$?
no — the sum should be $3$, not $-3$
yes — $(-1) + (-2) = -3$ and $(-1)(-2) = 2$
no — the product should be $-2$, since both zeros are negative
yes, but only by coincidence, since the method does not guarantee it
Check your answer
no — the sum should be $3$, not $-3$ — $(-1)+(-2)=-3$ exactly matches the original target sum; there is no sign error to correct here.
✓ yes — $(-1) + (-2) = -3$ and $(-1)(-2) = 2$ — (B) Adding and multiplying the two zeros directly reproduces exactly the target values the quadratic was built from.
no — the product should be $-2$, since both zeros are negative — A negative times a negative gives a POSITIVE result: $(-1)(-2)=2$, matching the target exactly.
yes, but only by coincidence, since the method does not guarantee it — The construction method GUARANTEES this match by design — $x^2-s x+p$ is built precisely so its zeros have sum $s$ and product $p$.
Four graphs meet the x-axis 2, 1, 0 and 4 times, giving each one’s count of zeros.
Worked example
Counting zeros from four graphs
graph (a) meets the x-axis at exactly two points Two distinct crossings give two distinct zeros.
graph (b) touches the x-axis at exactly one point, without crossing it A touch counts as one zero, not two and not zero. It is a single repeated zero.
graph (c) never meets the x-axis No crossing and no touch means no real zero at all.
graph (d) meets the x-axis at exactly four points The polynomial behind it must have degree at least $4$. A graph can never meet the x-axis more times than its degree.
marks on (a) 2, on (b) 1, on (c) 0, on (d) 4 Go back over the four graphs yourself. Put a mark wherever the curve meets the axis, then count the marks. The four numbers must match the counts just read off.
Count the zeros of a polynomial straight off its graph, with no algebra.
Count the crossings Count every point where the curve meets the x-axis. Each crossing is one zero.
Count a touch as one A curve that only touches the axis, without crossing to the other side, still counts as one zero, not two and not zero.
Match the count to the degree The crossing count is the smallest degree the polynomial can have. A graph never meets the x-axis more times than its degree.
Check yourself
A graph of $y=p(x)$ meets the x-axis at exactly two points. How many zeros does $p(x)$ have?
$1$
$4$
an unknown number, since the graph alone cannot say
$2$
Check your answer
$1$ — Two distinct crossing points each count as a separate zero, giving $2$ zeros, not $1$.
$4$ — The graph shows exactly two crossings; there is no reason to double that count to $4$.
an unknown number, since the graph alone cannot say — Counting crossing points directly IS a valid way to read off the number of zeros — that is the whole point of this method.
✓ $2$ — (D) Each distinct meeting point with the x-axis is a distinct zero, so two points give exactly two zeros.
A graph touches the x-axis at one point without crossing it. This is described as
$1$ repeated zero
$0$ zeros, since the curve does not cross
$2$ zeros at the same point
an undefined number of zeros
Check your answer
✓ $1$ repeated zero — (A) A single touch, without crossing, is exactly how a repeated zero appears on a graph.
$0$ zeros, since the curve does not cross — Touching the axis is still meeting it — that single touch is $1$ (repeated) zero, not $0$.
$2$ zeros at the same point — A repeated zero is counted as ONE zero appearing with multiplicity, not as two separate zeros.
an undefined number of zeros — This exact picture, a single touch, has a definite name and count: one repeated zero.
A graph of $y=p(x)$ never meets the x-axis at all. According to the method used to read zero counts from a graph, what can be said directly, without any algebra?
$p(x)$ has no real zero
$p(x)$ has exactly one zero, hidden from the graph
$p(x)$ must be a constant polynomial
the degree of $p(x)$ must be odd
Check your answer
✓ $p(x)$ has no real zero — (A) Never meeting the x-axis is exactly the graphical picture the chapter ties to having no real zero.
$p(x)$ has exactly one zero, hidden from the graph — If a zero existed, the graph would show it as a crossing or a touch; a graph with neither shows no real zero.
$p(x)$ must be a constant polynomial — Many non-constant polynomials, such as $x^2+1$, never meet the x-axis either; missing the axis does not require being constant.
the degree of $p(x)$ must be odd — An odd-degree polynomial is guaranteed to cross the x-axis at least once, so a graph with no crossing rules out an odd degree, not requires one.
The curve y equals x squared plus 3x plus 2 crosses the x-axis at minus 2 and minus 1, sum minus 3, product 2.
This one polynomial shows a zero, a graph point, a degree cap, and a sum and product, all together. Take $p(x) = 2 x^2 - 8 x + 6$.
- Zero: $p(k) = 0$. Here $p(1) = 0$.
- Graph: a zero is a point where $y = p(x)$ meets the x-axis. Here, $(1, 0)$ and $(3, 0)$.
- Degree cap: a quadratic has at most $2$ zeros. This one has exactly $2$, matching its degree.
- Sum and product: the zeros add to $-b/a$ and multiply to $c/a$. Here $a$ is $2$, $b$ is $-8$ and $c$ is $6$. So they add to $8/2$, which is $4$, and multiply to $6/2$, which is $3$. The zeros $1$ and $3$ fit both.
- Reverse: a target sum $-3$ and product $2$ build $x^2 + 3 x + 2$, whose zeros are $-1$ and $-2$.
We looked at one number, $k = 1$, three ways. It is an algebra fact and a graph point. It is also part of a sum and a product, both read off $a$, $b$, $c$.
Your turn: check the second zero, $3$, against the sum and product above. (Answer: $3$ is the other zero used in the sum $1 + 3 = 4$ and the product $1 \cdot 3 = 3$ already shown.)
Check yourself
This chapter’s zero-finding method (splitting the middle term) and its reverse (building a quadratic from a sum and product) are related because
they run the same relation in opposite directions
they use entirely different formulas with no shared relation
the reverse method only works when the zeros are irrational
splitting the middle term and building from sum-product always give the same final quadratic
Check your answer
✓ they run the same relation in opposite directions — (A) Both methods rest on the same sum-product relation between a quadratic and its zeros, used forward in one case and in reverse in the other.
they use entirely different formulas with no shared relation — Both methods rest on the SAME relation, $\alpha+\beta=-b/a$ and $\alpha \beta=c/a$ — one runs it forward, the other in reverse.
the reverse method only works when the zeros are irrational — The construction method works for any target sum and product, whether the resulting zeros turn out rational or irrational.
splitting the middle term and building from sum-product always give the same final quadratic — Splitting starts from a GIVEN quadratic; building from sum-product starts from a TARGET sum and product — these are different starting points, not the same one.
A quadratic’s zero, seen algebraically as $p(k)=0$ and seen graphically as a point on the x-axis, are
two separate properties that happen to coincide sometimes
the same fact about $p(x)$, described in two different ways
only equal when the polynomial is linear
different concepts, since one is a number and the other is a point
Check your answer
two separate properties that happen to coincide sometimes — The two descriptions ALWAYS agree, for every polynomial — they are not separate properties that merely happen to line up.
✓ the same fact about $p(x)$, described in two different ways — (B) The algebraic condition and the graphical picture are two descriptions of exactly the same underlying fact about the polynomial.
only equal when the polynomial is linear — This equivalence holds for a polynomial of any degree, not only linear ones — every worked example in this chapter uses it for quadratics.
different concepts, since one is a number and the other is a point — The number $k$ and the point $(k,0)$ describe the SAME location — one names its x-coordinate, the other its full position on the graph.
Given only that a quadratic’s zeros sum to $7$ and multiply to $10$, can you find the quadratic without factorising anything first?
no — the zeros must be factorised out of a quadratic first
yes, but only by guessing pairs of numbers that multiply to $10$
no — sum and product alone are never enough information
yes — build $x^2-7x+10$ directly, no factorising needed
Check your answer
no — the zeros must be factorised out of a quadratic first — The whole point of running the relation in reverse is to build the quadratic FIRST, directly from the sum and product, with no prior factorising.
yes, but only by guessing pairs of numbers that multiply to $10$ — $x^2-s x+p$ is a direct formula, not a guess — substituting $s=7,p=10$ gives the quadratic immediately.
no — sum and product alone are never enough information — Sum and product ARE enough: $x^2-s x+p$ is exactly the quadratic they determine, up to a nonzero scaling factor.
✓ yes — build $x^2-7x+10$ directly, no factorising needed — (D) The sum-product relation gives a direct formula for the quadratic from a target sum and product, with no factorising step needed first.
Check yourself: the whole chapter
A zero of a polynomial $p(x)$ is a value $k$ with $p(k) = 0$. On the graph of $y = p(x)$, what does a zero correspond to?
the value of $p(x)$ when $x = 0$
the highest point on the curve
the point where the curve crosses the y-axis
a point where the curve meets the x-axis
Check your answer
the value of $p(x)$ when $x = 0$ — $p(0)$ is the value of the polynomial when $x = 0$; a zero is a value of $x$ that makes $p(x)$ equal to $0$ — these are generally different numbers.
the highest point on the curve — The highest point of a curve is unrelated to where it crosses an axis — a zero is about crossing the x-axis, not about height.
the point where the curve crosses the y-axis — A zero is an x-axis crossing. Crossing the y-axis instead gives $p(0)$, a different value.
✓ a point where the curve meets the x-axis — (D) A zero of $p(x)$ is a value $k$ with $p(k) = 0$, and on the graph of $y = p(x)$ this is exactly a point where the curve meets the x-axis.
The graph of the linear polynomial $p(x) = 2x - 6$ is a straight line. Where does it meet the x-axis?
at $(0, -6)$
at $(3, 0)$
at $(0, 0)$
at $(-3, 0)$
Check your answer
at $(0, -6)$ — $(0, -6)$ is where the line meets the y-axis ($p(0) = -6$); the x-axis crossing needs $p(x) = 0$, solved for $x$.
✓ at $(3, 0)$ — (B) Solving $2x - 6 = 0$ gives $x = 3$, so the line meets the x-axis at $(3, 0)$ — the point where $p(x) = 0$.
at $(0, 0)$ — Nothing about $p(x) = 2x - 6$ places it through the origin — check by solving $2x - 6 = 0$ directly.
at $(-3, 0)$ — Solving $2x - 6 = 0$ gives $2x = 6$, so $x = 3$, not $x = -3$ — check the sign when moving $6$ across.
A polynomial has degree $3$. At most how many points can its graph meet the x-axis?
$6$, twice the degree
$1$, regardless of the degree
$3$, the same as the degree
as many as the constant term’s value
Check your answer
$6$, twice the degree — The cap on the number of zeros matches the degree directly — it is not doubled.
$1$, regardless of the degree — A polynomial of degree $3$ can have up to $3$ zeros, not always exactly $1$ — the degree sets the cap, not a fixed count.
✓ $3$, the same as the degree — (C) The graph of a degree-$n$ polynomial meets the x-axis at most $n$ times, so a degree-$3$ polynomial has at most $3$ zeros.
as many as the constant term’s value — The constant term plays no role in how many times the graph can cross the x-axis — the degree alone sets that cap.
For the quadratic $2x^2 - 7x + 3$, with zeros $\alpha$ and $\beta$, what are $\alpha + \beta$ and $\alpha \beta$?
$\alpha + \beta = -7/2$, $\alpha \beta = 3/2$
$\alpha + \beta = 7/2$, $\alpha \beta = 2/3$
$\alpha + \beta = -7$, $\alpha \beta = 3$
$\alpha + \beta = 7/2$, $\alpha \beta = 3/2$
Check your answer
$\alpha + \beta = -7/2$, $\alpha \beta = 3/2$ — The sum is $-b/a$, not $b/a$ — with $b = -7$ and $a = 2$, that is $-(-7)/2 = 7/2$, not $-7/2$.
$\alpha + \beta = 7/2$, $\alpha \beta = 2/3$ — The product is $c/a$, not $a/c$ — with $c = 3$ and $a = 2$, that is $3/2$, not $2/3$.
$\alpha + \beta = -7$, $\alpha \beta = 3$ — Both the sum and the product need dividing by $a$ — using $b$ and $c$ on their own only works when $a = 1$, and here $a = 2$.
To factorise $x^2 + 5x + 6$ by splitting the middle term, which two numbers should replace $5x$?
$6$ and $1$, since $6 \cdot 1 = 6$
$2$ and $3$, since $2 \cdot 3 = 6$ and $2 + 3 = 5$
$-2$ and $-3$, since $(-2) \cdot (-3) = 6$
$1$ and $5$, since $1 \cdot 5 = 5$ and $1 + 5 = 6$
Check your answer
$6$ and $1$, since $6 \cdot 1 = 6$ — $6$ and $1$ multiply to $6$, but they add to $7$, not $5$ — both conditions must hold together.
✓ $2$ and $3$, since $2 \cdot 3 = 6$ and $2 + 3 = 5$ — (B) $2$ and $3$ multiply to $6$ (which is $a \cdot c$) and add to $5$ (which is $b$), so splitting $5x$ into $2x + 3x$ factorises $x^2 + 5x + 6$.
$-2$ and $-3$, since $(-2) \cdot (-3) = 6$ — $-2$ and $-3$ do multiply to $6$, but they add to $-5$, not $5$ — the signs need to give a positive sum here.
$1$ and $5$, since $1 \cdot 5 = 5$ and $1 + 5 = 6$ — The two numbers must multiply to $a \cdot c = 6$ and add to $b = 5$ — not multiply to $b$ and add to $c$.
Between the two zeros the stall earns; outside them it loses, so the zeros are the prices to stay between.
You may never solve a polynomial for fun after Class 10. But shops price things using formulas just like these. Here are seven of those places.
Paying for a cab ride. Your cab charges a fixed $₹40$ to start. Then it charges $₹12$ for every km you ride. The maths: the fare for $x$ km is the value of that polynomial at $x$. For $x = 8$ km: $40 + 12 \cdot 8 = 136$ rupees.
Pricing a custom rug. A shop prices a square rug at $₹150$ for every square foot of its area. The maths: the price is degree $2$ in the side length, since area is the side squared. That makes it a quadratic. For a side of $6$ feet: $150 \cdot 6^2 = 5400$ rupees.
Reading a stall’s profit chart. A stall’s daily profit is plotted against the price of a plate. The curve crosses the price axis at two points. The maths: each point where the curve crosses the axis is a zero of the profit polynomial. Two zeros here means two prices give zero profit. Profit is $-(x - 7)(x - 13)$ rupees, for a plate priced at $x$ rupees. It is $0$ at $x = 7$ and $x = 13$.
A taxi meter runs a degree-one polynomial, and every fare it shows is one value of that polynomial.
Finding a lemonade stand’s break-even prices. Your lemonade stand’s daily profit is $-x^2 + 34 x - 240$ rupees, for a glass priced at $x$ rupees. The maths: splitting the middle term finds the exact prices where profit is $0$. Split $34$ into $10$ and $24$: $10 \cdot 24 = 240$ and $10 + 24 = 34$. So profit is $0$ at $x = 10$ and $x = 24$.
Checking a stall’s break-even claim. A friend says their samosa stall breaks even at $₹9$ and $₹15$ a plate. Their profit formula is $-x^2 + 24 x - 135$. The maths: the sum and product of the zeros come straight from the coefficients, with no solving. Sum $= 24$ and product $= 135$. And $9 + 15 = 24$, $9 \cdot 15 = 135$. It matches.
Writing a vendor’s profit formula. A vendor breaks even selling $3$ kg or $12$ kg of guavas a day. That is all you know about the profit formula. The maths: knowing just the sum and product of the zeros builds the whole quadratic shape. Sum $= 3 + 12 = 15$ and product $= 3 \cdot 12 = 36$. So the profit is a multiple of $x^2 - 15 x + 36$, which is $0$ at $x = 3$ and $x = 12$.
Planning a rectangular garden. You have $40$ m of fence for a rectangular garden. If one side is $x$ m, the other is $20 - x$ m. The maths: the area $20 x - x^2$ has degree $2$, so it has at most two zeros. The zeros are $x = 0$ and $x = 20$, the two widths with no garden at all. At $x = 10$ the area is $100$ square m.
Your turn. A juice stall’s profit is $-x^2 + 16 x - 39$ rupees, for a glass priced at $x$ rupees. At what two prices does it break even? Answer: Split $16$ into $3$ and $13$: $3 \cdot 13 = 39$ and $3 + 13 = 16$. So it breaks even at $x = 3$ and $x = 13$ rupees.
practice A graph of $y = p(x)$ meets the x-axis at $x = -2$ and at $x = 3$, and nowhere else. How many zeros does $p(x)$ have, and what are they? (Worked in full below — read it, then do the next two the same way.)
practice A graph of $y = p(x)$ meets the x-axis at $x = -1$, at $x = 0$ and at $x = 5$, and nowhere else. How many zeros does $p(x)$ have? (Same first step — list the x-values where the graph meets the axis, then count them.)
practice A graph of $y = p(x)$ touches the x-axis at $x = 2$ without crossing it there, and crosses the axis at $x = -4$. How many zeros does $p(x)$ have, counted with repetition? (A touch point counts twice.)
practice A graph of $y = p(x)$ crosses the x-axis at exactly two points and does not touch it anywhere else. How many zeros does $p(x)$ have?
practice A graph of $y = p(x)$ touches the x-axis at exactly one point, without crossing it, and meets the axis nowhere else. How many zeros does $p(x)$ have, counted with repetition?
practice A graph of $y = p(x)$ does not meet the x-axis at all. How many real zeros does $p(x)$ have?
practice A graph of $y = p(x)$ meets the x-axis at exactly four points. What is the least possible degree of $p(x)$?
practice Can the graph of a quadratic polynomial meet the x-axis at three distinct points? Give a reason.
Answers
$2$ zeros — $-2$ and $3$.
$3$ zeros — $-1$, $0$ and $5$.
$3$ zeros counted with repetition — the touch at $x = 2$ counts twice, and the crossing at $x = -4$ counts once.
2 zeros.
$2$ zeros counted with repetition — the touch point is a double root, so there is one distinct zero, counted twice.
No real zero.
Degree at least 4.
No — a quadratic’s graph meets the x-axis in at most 2 points, since its degree caps the zero count at 2.
Exercise 2.1 — further practice, counting zeros
practice The graph of $y = p(x)$ comes down from the upper left, reaches its lowest point below the x-axis, and then rises to the upper right, crossing the x-axis at exactly two points along the way. It meets the axis nowhere else. How many zeros does $p(x)$ have?
practice The graph of $y = p(x)$ crosses the x-axis at exactly three distinct points and meets the axis nowhere else. How many zeros does $p(x)$ have?
practice The graph of $y = p(x)$ touches the x-axis at two different points, without crossing the axis at either one, and meets the axis nowhere else. How many zeros does $p(x)$ have?
practice The graph of $y = p(x)$ touches the x-axis at exactly one point without crossing it there, and crosses the axis at two further points. It meets the axis nowhere else. How many zeros does $p(x)$ have?
$1$
$2$
$3$
$4$
practice The graph of $y = p(x)$ meets the x-axis at exactly six distinct points and meets it nowhere else. What is the least possible degree of $p(x)$?
practice A quadratic polynomial’s graph opens upward, and its lowest point lies exactly on the x-axis. How many zeros does the polynomial have, and are the two zeros equal or different?
Answers
$2$ zeros.
$3$ zeros.
$2$ zeros (each a repeated zero).
C — $3$ zeros (the touch point is a repeated zero).
Degree at least $6$.
$2$ equal zeros — the lowest point sitting on the x-axis makes it a double root, so the two zeros coincide at that one point.
practice Find the zeros of $2 x^2 - 10 x + 12$, and verify the relation between the zeros and the coefficients. (Worked in full below — it is the worked example with new numbers.)
practice Find the zeros of $3 x^2 - 12 x + 9$, and verify the relation between the zeros and the coefficients. (Same first step — take the common factor $3$ out, then split the middle term. Keep $a = 3$ in the check.)
practice Find a quadratic polynomial whose zeros have sum $-4$ and product $-5$. (Use $x^2 - s x + p$, with $s$ the sum and $p$ the product.)
practice Find the zeros of $x^2 - 2 x - 8$, and verify the relation between the zeros and the coefficients.
practice Find the zeros of $6 x^2 - 7 x - 3$, and verify the relation between the zeros and the coefficients.
practice Find the zeros of $4 x^2 - 4 x + 1$, and verify the relation between the zeros and the coefficients.
practice Find a quadratic polynomial whose zeros are $1/4$ and $-1$.
practice Find a quadratic polynomial whose sum and product of zeros are $5$ and $6$ respectively.
Answers
Zeros $2$ and $3$; sum $5 = -(-10)/2$, product $6 = 12/2$.
Zeros $1$ and $3$; sum $4 = -(-12)/3$, product $3 = 9/3$.
$x^2 + 4 x - 5$ (or any nonzero multiple).
Zeros $-2$ and $4$; sum $2 = -(-2)/1$, product $-8 = -8/1$.
Zeros $-1/3$ and $3/2$; sum $7/6 = -(-7)/6$, product $-1/2 = -3/6$.
Zeros $1/2$ and $1/2$ (repeated); sum $1 = -(-4)/4$, product $1/4 = 1/4$.
$4 x^2 + 3 x - 1$ (or any nonzero multiple).
$x^2 - 5 x + 6$.
Exercise 2.2 — further practice, zeros and coefficients
practice Find the zeros of $x^2 - 7 x + 12$, and verify the relation between the zeros and the coefficients.
practice Find the zeros of $x^2 + x - 12$, and verify the relation between the zeros and the coefficients.
practice Find the zeros of $5 u^2 + 10 u$, and verify the relation between the zeros and the coefficients.
practice Find a quadratic polynomial whose zeros have sum $-2$ and product $-15$.
practice For the quadratic $2 x^2 - 3 x - 5$ with zeros $\alpha$ and $\beta$, what is $\alpha + \beta$?
$3/2$
$-3/2$
$5/2$
$-5/2$
practice Find the zeros of $2 x^2 - 5 x + 2$, and verify the relation between the zeros and the coefficients.
practice Find a quadratic polynomial whose zeros are $2/3$ and $-3$.
practice Which of these is a quadratic polynomial whose zeros have sum $0$ and product $-9$?
$x^2 - 9$
$x^2 + 9$
$x^2 - 9 x$
$x^2 + 9 x$
practice Find the zeros of $x^2 - 20$, and verify the relation between the zeros and the coefficients.
practice Find the zeros of $-x^2 + x + 6$, and verify the relation between the zeros and the coefficients.
Answers
Zeros $3$ and $4$; sum $7 = -(-7)/1$, product $12 = 12/1$.
Zeros $3$ and $-4$; sum $-1 = -1/1$, product $-12 = -12/1$.
Zeros $0$ and $-2$; sum $-2 = -10/5$, product $0 = 0/5$.
$x^2 + 2 x - 15$ (or any nonzero multiple).
A — $\alpha + \beta = 3/2$.
Zeros $1/2$ and $2$; sum $5/2 = -(-5)/2$, product $1 = 2/2$.
$3 x^2 + 7 x - 6$ (or any nonzero multiple).
A — $x^2 - 9$.
Zeros $2 \sqrt{5}$ and $-2 \sqrt{5}$; sum $0 = -0/1$, product $-20 = -20/1$.
Zeros $3$ and $-2$; sum $1 = -1/(-1)$, product $-6 = 6/(-1)$.
