Polynomial Zeros and Graph Features, SAT math lesson

Suppose a question shows you $p(x) = (x - 4)(x + 1)^2$ and asks where the graph meets the x-axis. You could expand it and hunt for roots, but the factored form already holds the answer. A product is zero only when one of its factors is zero, so the graph meets the axis at $x = 4$ and $x = -1$ . The factors also tell you something subtler, namely whether the graph passes through the axis there or only bounces off it.

See also Polynomial Zeros, Signs, and Graph Shape in Foundations for Calculus and Zeros and Multiplicity of Polynomials in Precalculus.

Key terms

Zero of a polynomial

Definition A polynomial is a function of the form

\begin{aligned} p(x) = a_n x^n &+ a_{n-1} x^{n-1} + \dots \\[0.4em] &+ a_1 x + a_0, \end{aligned}

where $n$ is a nonnegative integer and the coefficients are real numbers. If $a_n \ne 0$ , the degree is $n$ and $a_n$ is the leading coefficient. A number $r$ is a zero (or root) of $p$ when $p(r) = 0$ . On the SAT, unless a question says otherwise, only real zeros count, because only they appear on a graph. The key fact for factored forms is the zero product property. A product of real numbers is zero exactly when at least one factor is zero.

In plain words A zero is an input that makes the output exactly $0$ . In factored form you find the zeros by setting each factor equal to $0$ on its own.

Example Take

p(x) = (x - 2)(x + 3)(2x - 1).

Setting each factor to zero gives $x = 2$ , $x = -3$ and $x = \frac{1}{2}$ . These are all the zeros. Notice that the factor $2x - 1$ gives $\frac{1}{2}$ , not $1$ , a detail that test writers like to use in wrong answer choices.

Factor and the factor theorem

Definition A polynomial $d(x)$ is a factor of $p(x)$ when $p(x) = d(x) \cdot q(x)$ for some polynomial $q(x)$ . The factor theorem says that for a polynomial $p$ and a real number $r$ ,

\begin{gathered} p(r) = 0 \quad \text{exactly when} \\[0.4em] (x - r) \text{ is a factor of } p(x). \end{gathered}

So every real zero $r$ matches a linear factor $x - r$ , and every linear factor $ax - b$ with $a \ne 0$ matches the zero $\frac{b}{a}$ .

In plain words Zeros and linear factors are two descriptions of the same thing. If you know one, you know the other, with the sign switched.

Example Let $p(x) = x^3 - 2x^2 - 5x + 6$ . Testing $x = 1$ gives $1 - 2 - 5 + 6 = 0$ , so $x - 1$ is a factor. Dividing it out leaves a quadratic that factors further,

\begin{aligned} p(x) &= (x - 1)(x^2 - x - 6) \\[0.4em] &= (x - 1)(x - 3)(x + 2). \end{aligned}

The zeros are $1$ , $3$ and $-2$ .

Structure often does this work faster than testing values. Grouping gives

\begin{aligned} &x^3 + 2x^2 - 9x - 18 \\[0.4em] &= x^2(x + 2) - 9(x + 2) \\[0.4em] &= (x + 2)(x - 3)(x + 3), \end{aligned}

with real zeros $-2$ , $3$ and $-3$ . A factor with no real zeros adds nothing to the graph’s crossings. For example,

x^4 - 16 = (x^2 + 4)(x - 2)(x + 2),

and since $x^2 + 4 \ge 4$ for every real $x$ , the only real zeros are $2$ and $-2$ .

Intercepts

Definition The x-intercepts of the graph of $y = p(x)$ are the points $(r, 0)$ where $p(r) = 0$ , so they sit exactly at the real zeros. The y-intercept is the point $(0, p(0))$ . A polynomial is defined for every real number, so its graph always has exactly one y-intercept, and its y-coordinate is the constant term $a_0$ .

In plain words The x-intercepts are where the graph meets the horizontal axis, and the zeros are their x-coordinates. The y-intercept is where you land when you plug in $0$ .

Example For $f(x) = (x + 1)(x - 2)(x - 4)$ the x-intercepts are $(-1, 0)$ , $(2, 0)$ and $(4, 0)$ . The y-intercept comes from

f(0) = (1)(-2)(-4) = 8,

so it is $(0, 8)$ .

Sign of a polynomial between zeros

Definition A polynomial is continuous, so it can change sign only at a real zero. Between two consecutive real zeros, and also to the left of the smallest one and to the right of the largest one, its sign stays the same. Its sign on each interval is the product of the signs of its factors there. A factor $x - r$ is negative for $x < r$ and positive for $x > r$ , an even power $(x - r)^2$ is positive for every $x \ne r$ , and a factor with no real zeros, such as $x^2 + 4$ , keeps one sign everywhere.

In plain words The real zeros cut the number line into pieces. On each piece the graph stays entirely above the axis or entirely below it, and one test input tells you which.

Example Let $g(x) = (x + 2)(x - 1)^2$ . The zeros $-2$ and $1$ split the line into three intervals. Test one input in each.

Interval	Test input	$x + 2$	$(x - 1)^2$	Sign of $g$
$x < -2$	$-3$	$-$	$+$	negative, $g(-3) = -16$
$-2 < x < 1$	$0$	$+$	$+$	positive, $g(0) = 2$
$x > 1$	$2$	$+$	$+$	positive, $g(2) = 4$

The sign changes at $-2$ but not at $1$ . That difference is the subject of the next term.

Multiplicity

Definition A real zero $r$ of $p$ has multiplicity $k$ when $(x - r)^k$ is a factor of $p(x)$ but $(x - r)^{k+1}$ is not. Multiplicity describes the graph only near that zero.

If $k$ is odd, $p(x)$ changes sign at $r$ , so the graph crosses the x-axis there. When $k = 1$ it crosses at an angle, like a line. When $k$ is odd and at least $3$ , it flattens out against the axis as it crosses.
If $k$ is even, $p(x)$ keeps the same sign on both sides of $r$ , so the graph touches the x-axis and turns back without crossing. The larger the even multiplicity, the flatter the graph looks near $r$ .

In plain words Count how many times a factor repeats. An odd count means the graph goes through the axis, an even count means it touches and bounces back. Higher counts only make the graph hug the axis more closely near that zero.

Example For $g(x) = (x + 2)(x - 1)^2$ the zero $-2$ has multiplicity $1$ and the graph crosses there, from negative to positive. The zero $1$ has multiplicity $2$ and the graph touches the axis at $(1, 0)$ while staying positive on both sides, as the sign table showed. Multiplicity says nothing about how high the graph rises between $-2$ and $1$ . For that you need actual values, such as $g(0) = 2$ .

Polynomial graph

Definition The graph of a polynomial is a single smooth, unbroken curve with no holes, jumps or asymptotes. A polynomial of degree $n \ge 1$ has at most $n$ real zeros, counted with multiplicity. Its end behavior is set by the leading term $a_n x^n$ . For even $n$ both ends point the same way, up when $a_n > 0$ and down when $a_n < 0$ . For odd $n$ the ends point opposite ways. When $a_n > 0$ the graph falls on the left and rises on the right, and when $a_n < 0$ it does the reverse.

Two cautions follow. The zeros alone do not determine the graph, since $(x - 1)(x - 3)$ , $-2(x - 1)(x - 3)$ and $(x - 1)(x - 3)(x^2 + 1)$ all have the zeros $1$ and $3$ but different shapes and heights. And a graph on a screen shows only a window, so zeros far away can be hidden.

In plain words The factors tell you where the graph meets the axis and how it behaves right there. To pin down the rest you need extra information, such as the leading coefficient, the degree or one more point.

Example A graph crosses the x-axis only at $x = -1$ and $x = 3$ , each with multiplicity $1$ , and passes through $(0, 6)$ . Assume it is a quadratic. Then

f(x) = a(x + 1)(x - 3)

for some constant $a \ne 0$ . The point gives $f(0) = -3a = 6$ , so $a = -2$ and

f(x) = -2(x + 1)(x - 3).

Without the point, every nonzero $a$ would fit the zeros.

Now the window warning. The cubic

h(x) = (x + 1)(x - 1)(x - 50)

has $h(x) < 0$ for every $x$ between $1$ and $50$ . On a window from $x = -10$ to $x = 10$ it shows only the two zeros $-1$ and $1$ , and both ends of the visible curve point down, so it looks like a quadratic opening downward. The third zero at $50$ and the rise on the far right are simply off screen. Its leading coefficient is $1$ , and $h(x) > 0$ for every $x > 50$ . In the Desmos graphing calculator, zoom out or check the factored form before you trust a count of zeros from a picture.

Check yourself

The function $f$ is defined by $f(x) = 2(x - 4)(x + 1)^2$ . At which value of $x$ does the graph of $y = f(x)$ touch the x-axis without crossing it? (A) $-4$ (B) $-1$ (C) $1$ (D) $4$

Show the answer

The answer is (B). The factor $x + 1$ appears twice, so $-1$ is a zero of even multiplicity and $f$ keeps its sign there. Choice (D) is a zero, but of multiplicity $1$ , so the graph crosses at $4$ . Choices (A) and (C) come from reading the signs in the factors backwards, and neither makes $f$ equal to $0$ .

The graph of $y = a x(x - 2)(x + 3)$ , where $a$ is a constant, passes through the point $(1, 12)$ . What is the value of $a$ ?

Show the answer

Substituting $x = 1$ and $y = 12$ gives

12 = a(1)(1 - 2)(1 + 3) = -4a,

so $a = -3$ . The tempting slip is to drop the sign of $1 - 2$ and get $3$ .

How many distinct real zeros does $p(x) = x^4 - 3x^2 - 4$ have? (A) $0$ (B) $2$ (C) $3$ (D) $4$

Show the answer

The answer is (B). Treat $x^2$ as one quantity and factor,

x^4 - 3x^2 - 4 = (x^2 - 4)(x^2 + 1).

The factor $x^2 - 4$ gives the real zeros $2$ and $-2$ . The factor $x^2 + 1$ is at least $1$ for every real $x$ , so it adds no real zeros. Choice (D) counts the degree, which is only an upper limit, and choice (C) adds a zero for the factor $x^2 + 1$ as if it gave one real solution, but $x^2 + 1 = 0$ has none. Choice (A) comes from seeing that $x^2 + 1$ has no real zeros and wrongly concluding the same for all of $p$ .

Polynomial Zeros and Graph Features

Theory, part 57

Practice, part 58

Key terms

Zero of a polynomial

Factor and the factor theorem

Intercepts

Sign of a polynomial between zeros

Multiplicity

Polynomial graph

Check yourself

How this lesson connects

Builds on

Used later in