Zeros and Multiplicity of Polynomials, precalculus lesson

A factored polynomial such as $p(x) = (x + 2)(x - 1)^2$ tells you a lot before you plot a single point. The factors give the places where the graph meets the $x$ -axis, and the exponents tell you how it meets the axis there, by crossing straight through, by touching and turning back, or by flattening out as it passes.

Key terms

Zero

Definition A zero of a polynomial $p$ is a number $c$ with $p(c) = 0$ . A zero can be a real number or a nonreal complex number, and a real zero is a zero that is a real number.

In plain words A zero is an input that makes the output exactly $0$ .

Example For $p(x) = x^2 - 5x + 6 = (x - 2)(x - 3)$ the zeros are $2$ and $3$ . Indeed $p(2) = 4 - 10 + 6 = 0$ and $p(3) = 9 - 15 + 6 = 0$ .

Root

Definition A root of the equation $p(x) = 0$ is a solution of that equation. The roots of $p(x) = 0$ are exactly the zeros of $p$ , and many texts also call them the roots of $p$ . When $p$ is written as a product of polynomial factors, the zero product property says that $p(c) = 0$ exactly when at least one factor is $0$ at $c$ .

In plain words The word zero belongs to the function and the word root belongs to the equation, but they name the same numbers. Factoring turns the search for roots into several small equations.

Example To find the roots of $x^3 = 4x$ , move everything to one side and factor.

x^3 - 4x = x(x - 2)(x + 2) = 0.

The roots are $-2$ , $0$ and $2$ . Dividing both sides of $x^3 = 4x$ by $x$ would have lost the root $0$ .

x-intercept

Definition An $x$ -intercept of the graph of $p$ is a point where the graph meets the $x$ -axis. It has the form $(c, 0)$ where $c$ is a real zero of $p$ , and every real zero gives one. Nonreal zeros give no $x$ -intercepts, because every point of the graph has real coordinates.

In plain words The $x$ -intercepts are the real zeros, seen on the graph. Many texts give only the number $c$ and call that the intercept.

Example The polynomial $p(x) = (x - 1)(x^2 + 4)$ has only one $x$ -intercept, the point $(1, 0)$ . The factor $x^2 + 4$ is at least $4$ for every real $x$ , so it is never $0$ on the real line.

Multiplicity

Definition Let $c$ be a zero of a nonzero polynomial $p$ . The multiplicity of $c$ is the positive integer $m$ such that

p(x) = (x - c)^m \, q(x)

for some polynomial $q$ with $q(c) \ne 0$ . This $m$ always exists and is unique. A zero of multiplicity $1$ is a simple zero and a zero of multiplicity $2$ is a double zero. The multiplicities of the distinct zeros of $p$ add up to at most the degree of $p$ .

In plain words The multiplicity counts how many times the factor $x - c$ appears once the polynomial is factored completely.

Example In $p(x) = (x - 3)^2 (x^2 - 9)$ it is tempting to say that $3$ has multiplicity $2$ . But $x^2 - 9 = (x - 3)(x + 3)$ , so

p(x) = (x - 3)^3 (x + 3).

The zero $3$ has multiplicity $3$ and the zero $-3$ has multiplicity $1$ . The total $3 + 1 = 4$ equals the degree.

Odd multiplicity and sign change

Definition Let $p$ have real coefficients and let $c$ be a real zero of multiplicity $m$ , so $p(x) = (x - c)^m q(x)$ with $q(c) \ne 0$ . A polynomial has no jumps, so $q$ keeps the sign of $q(c)$ on some open interval around $c$ . If $m$ is odd, the factor $(x - c)^m$ is negative for $x < c$ and positive for $x > c$ , so on that interval $p(x)$ has one sign to the left of $c$ and the opposite sign to the right. The graph crosses the $x$ -axis at $(c, 0)$ .

In plain words An odd power of $x - c$ flips sign as $x$ passes $c$ , and the rest of the polynomial does not, so the graph passes from one side of the axis to the other.

Example For $p(x) = (x - 1)^3 (x + 2)$ the zero $1$ has multiplicity $3$ . Just to either side of it,

\begin{aligned} p(0.9) &= (-0.1)^3 (2.9) = -0.0029, \\[0.4em] p(1.1) &= (0.1)^3 (3.1) = 0.0031. \end{aligned}

The sign goes from negative to positive, so the graph crosses the axis at $(1, 0)$ .

Even multiplicity and no sign change

Definition With the same notation, if $m$ is even, then $(x - c)^m > 0$ for every $x \ne c$ . So near $c$ the values $p(x)$ have the sign of $q(c)$ on both sides of $c$ , and there is no sign change at $c$ . The graph touches the $x$ -axis at $(c, 0)$ and turns back to the side it came from.

In plain words An even power is never negative, so the graph comes down to the axis (or up to it), touches it, and goes back.

Example A table of signs for $p(x) = (x + 2)(x - 1)^2$ uses one test point in each of the open intervals into which the real zeros cut the number line, since a polynomial keeps one sign on each such interval.

Interval	$(-\infty, -2)$	$(-2, 1)$	$(1, \infty)$
Test point	$-3$	$0$	$2$
Value of $p$	$-16$	$2$	$4$
Sign	negative	positive	positive

The sign changes at the simple zero $-2$ and stays positive across the double zero $1$ . So the graph crosses at $(-2, 0)$ and touches at $(1, 0)$ .

Local flattening

Definition Near a real zero $c$ of multiplicity $m$ , with $p(x) = (x - c)^m q(x)$ and $q(c) \ne 0$ , the polynomial behaves like $q(c)(x - c)^m$ in the sense that

\begin{gathered} \frac{p(x)}{q(c)(x - c)^m} = \frac{q(x)}{q(c)} \to 1 \\[0.8em] \text{as } x \to c. \end{gathered}

The secant line from $(c, 0)$ to a nearby point $(x, p(x))$ of the graph has slope

\begin{gathered} \frac{p(x) - 0}{x - c} = (x - c)^{m - 1} q(x), \\[0.8em] x \ne c. \end{gathered}

If $m = 1$ this slope tends to $q(c) \ne 0$ , and the graph cuts the axis at a slant. If $m \ge 2$ it tends to $0$ , and the graph meets the axis flat. The larger $m$ is, the faster $(x - c)^m$ shrinks as $x$ approaches $c$ , so the flatter the graph is there.

In plain words Close to a zero, the graph looks like a stretched copy of $y = (x - c)^m$ . A simple zero looks like a slanted line through the axis, a double zero like a parabola that touches it from above or below, and a triple zero like the curve $y = x^3$ , which levels off as it crosses.

Example At $x = 1.1$ , a distance of $0.1$ from the zero $c = 1$ , compare the powers of $x - 1$ .

$m$	$(x - 1)^m$ at $x = 1.1$	Secant slope from $(1, 0)$
$1$	$0.1$	$1$
$2$	$0.01$	$0.1$
$3$	$0.001$	$0.01$
$4$	$0.0001$	$0.001$

Each extra factor of $x - 1$ shrinks the value by a factor of $10$ at this distance. For $p(x) = (x - 1)^3 (x + 2)$ from the sign change example, $q(x) = x + 2$ and $q(1) = 3$ , so near $x = 1$ the graph looks like $y = 3(x - 1)^3$ . At $x = 1.1$ that gives $0.003$ , close to the true value $0.0031$ .

Real and nonreal roots

Definition A polynomial with real coefficients can have zeros that are not real. These are complex numbers $a + bi$ with $a$ and $b$ real, $b \ne 0$ and $i^2 = -1$ , which get their own lesson later in this unit. They come in conjugate pairs. If $a + bi$ is a zero of multiplicity $m$ , then its conjugate $a - bi$ is also a zero of multiplicity $m$ . A later lesson shows that a polynomial of degree $n \ge 1$ has exactly $n$ complex zeros counted with multiplicity. So the number of real zeros, counted with multiplicity, is $n$ minus an even number, and every polynomial with real coefficients and odd degree has at least one real zero. Only the real zeros appear as $x$ -intercepts.

In plain words A quadratic factor with no real zeros, such as $x^2 + 1$ , still has two zeros. They lie off the real number line, so the graph never shows them.

Example The polynomial $x^4 - 1$ factors as

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

Its real zeros $1$ and $-1$ give the $x$ -intercepts. The factor $x^2 + 1$ has discriminant $0^2 - 4 \cdot 1 \cdot 1 = -4 < 0$ , and its zeros are the nonreal numbers $i$ and $-i$ , since $i^2 + 1 = 0$ . That makes $4$ zeros in all, matching the degree.

Check yourself

List the zeros of $p(x) = x^3 (x + 4)^2 (x - 5)$ with their multiplicities. Where does the graph cross the $x$ -axis and where does it only touch it?

Show the answer

The zero $0$ has multiplicity $3$ , so the graph crosses at the origin and flattens as it crosses. The zero $-4$ has multiplicity $2$ , so the graph touches the axis at $(-4, 0)$ and turns back. The zero $5$ has multiplicity $1$ , so the graph crosses at $(5, 0)$ at a slant.

Find all zeros of $p(x) = x^4 + 3x^2 - 4$ and give the $x$ -intercepts of its graph.

Show the answer

As a quadratic in $x^2$ it factors as

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

The real zeros are $1$ and $-1$ , so the $x$ -intercepts are $(-1, 0)$ and $(1, 0)$ . The factor $x^2 + 4$ gives the nonreal zeros $2i$ and $-2i$ .

Find the polynomial of degree $4$ with leading coefficient $1$ whose graph crosses the $x$ -axis at $x = -1$ and $x = 0$ and touches it at $x = 3$ . Why is there only one such polynomial?

Show the answer

Crossing needs odd multiplicity, so at least $1$ at each of $-1$ and $0$ . Touching needs even multiplicity, so at least $2$ at $3$ . That already adds up to $1 + 1 + 2 = 4$ , the degree, so these multiplicities are exact and there are no other zeros. With leading coefficient $1$ the polynomial is

\begin{aligned} p(x) &= x(x + 1)(x - 3)^2 \\[0.4em] &= x^4 - 5x^3 + 3x^2 + 9x. \end{aligned}

Zeros and Multiplicity of Polynomials

Key terms

Zero

Root

x-intercept

Multiplicity

Odd multiplicity and sign change

Even multiplicity and no sign change

Local flattening

Real and nonreal roots

Check yourself

How this lesson connects

Builds on

Used later in

Where this leads in Calculus