Math Core

A graph shows you a polynomial's real zeros as $x$ -intercepts, but it can't give you exact values, and it hides the complex zeros completely. This lesson collects the algebra tools that find every zero exactly: the rational zero theorem to generate candidates, synthetic division to test them and shrink the polynomial, Descartes' rule of signs to know what to expect, and the fundamental theorem of algebra to know when you're done.

The rational zero theorem

Guessing zeros at random is hopeless. Fortunately, for polynomials with integer coefficients, any rational zero must come from a short list.

Rational zero theorem

If $f(x) = a_n x^n + \cdots + a_1 x + a_0$ has integer coefficients and $\dfrac{p}{q}$ is a rational zero in lowest terms, then

$p$ is a factor of the constant term $a_0$ , and
$q$ is a factor of the leading coefficient $a_n$ .

Here's why. If $f\left(\dfrac{p}{q}\right) = 0$ , multiply through by $q^n$ :

a_n p^n + a_{n-1}p^{n-1}q + \cdots + a_1 p q^{n-1} + a_0 q^n = 0.

Every term except the last contains a factor of $p$ , so $p$ must divide $a_0 q^n$ . Since $p$ and $q$ share no factors, $p$ divides $a_0$ . The same argument with the first term shows $q$ divides $a_n$ .

The theorem only lists candidates. You still have to test them, and a polynomial may have no rational zeros at all.

Testing candidates with synthetic division

Synthetic division by $x - c$ does two jobs at once. The last number is the remainder, which equals $f(c)$ , so a remainder of $0$ means $c$ is a zero. The other numbers are the coefficients of the quotient, a polynomial one degree lower, called the depressed polynomial. Keep dividing until you reach a quadratic, then finish with factoring or the quadratic formula.

Worked example: Finding all rational zeros

Find all zeros of $f(x) = 2x^3 - 3x^2 - 11x + 6$ .

Candidates. Factors of $6$ : $\pm 1, \pm 2, \pm 3, \pm 6$ . Factors of $2$ : $\pm 1, \pm 2$ . So the possible rational zeros are

\pm 1, \ \pm 2, \ \pm 3, \ \pm 6, \ \pm \tfrac{1}{2}, \ \pm \tfrac{3}{2}.

Test. $f(1) = 2 - 3 - 11 + 6 = -6$ , so $1$ is not a zero. Try $3$ with synthetic division:

	$2$	$-3$	$-11$	$6$
$3$		$6$	$9$	$-6$
	$2$	$3$	$-2$	$0$

The remainder is $0$ , so $3$ is a zero and $f(x) = (x - 3)(2x^2 + 3x - 2)$ .

Finish. $2x^2 + 3x - 2 = (2x - 1)(x + 2)$ . The zeros are $3$ , $\dfrac{1}{2}$ and $-2$ .

Irrational zeros

Once the depressed polynomial is a quadratic, the quadratic formula finds its zeros whether or not they are rational. Irrational zeros of a polynomial with rational coefficients come in pairs $a \pm \sqrt{b}$ , just like the quadratic formula produces them.

Worked example: A rational zero and an irrational pair

Find all zeros of $f(x) = x^3 - 3x^2 - 2x + 4$ .

The leading coefficient is $1$ , so the candidates are just the factors of $4$ : $\pm 1, \pm 2, \pm 4$ . Try $1$ : $f(1) = 1 - 3 - 2 + 4 = 0$ . ✓

	$1$	$-3$	$-2$	$4$
$1$		$1$	$-2$	$-4$
	$1$	$-2$	$-4$	$0$

The depressed polynomial is $x^2 - 2x - 4$ , which doesn't factor over the integers. Use the quadratic formula:

x = \frac{2 \pm \sqrt{4 + 16}}{2} = \frac{2 \pm 2\sqrt{5}}{2} = 1 \pm \sqrt{5}.

The zeros are $1$ , $1 + \sqrt{5}$ and $1 - \sqrt{5}$ .

Descartes' rule of signs

Before you start testing, it helps to know how many positive and negative zeros to expect. Write $f(x)$ in descending order and count the sign changes between consecutive nonzero coefficients.

Descartes' rule of signs

Let $f$ have real coefficients.

The number of positive real zeros equals the number of sign changes in $f(x)$ , or is less than that by an even number.
The number of negative real zeros equals the number of sign changes in $f(-x)$ , or is less than that by an even number.

Zeros are counted with multiplicity.

"Less by an even number" is because non-real zeros come in conjugate pairs, so real zeros disappear two at a time.

Worked example: Counting sign changes

What does Descartes' rule say about $f(x) = x^4 - 2x^3 + x^2 + 2x - 6$ ?

The signs of the coefficients are $+ \, - \, + \, + \, -$ . The changes are $+$ to $-$ , $-$ to $+$ , and $+$ to $-$ : three changes. So $f$ has $3$ or $1$ positive real zeros.

For negative zeros, replace $x$ with $-x$ . Even powers keep their sign and odd powers flip:

f(-x) = x^4 + 2x^3 + x^2 - 2x - 6.

The signs are $+ \, + \, + \, - \, -$ : exactly one change. So $f$ has exactly $1$ negative real zero. That's worth knowing: once you find one negative zero, stop testing negative candidates.

Complex zeros and complete factorization

The fundamental theorem of algebra guarantees that a polynomial of degree $n$ has exactly $n$ complex zeros, counted with multiplicity. Combined with the factor theorem, that means every polynomial splits completely into linear factors.

Linear factorization theorem

A polynomial of degree $n \ge 1$ with leading coefficient $a_n$ can be written as

f(x) = a_n(x - c_1)(x - c_2)\cdots(x - c_n),

where $c_1, \ldots, c_n$ are its complex zeros (some may repeat). If the coefficients are real, the non-real zeros come in conjugate pairs $a \pm bi$ .

So your search is finished when you've found $n$ zeros. Over the reals, each conjugate pair stays together as an irreducible quadratic factor.

Worked example: Real and non-real zeros

Find all zeros of $f(x) = x^4 - x^3 + 2x^2 - 4x - 8$ , and write $f$ as a product of linear factors.

Candidates: $\pm 1, \pm 2, \pm 4, \pm 8$ . Test: $f(2) = 16 - 8 + 8 - 8 - 8 = 0$ ✓ and $f(-1) = 1 + 1 + 2 + 4 - 8 = 0$ ✓. So $(x - 2)(x + 1) = x^2 - x - 2$ is a factor. Divide (or use synthetic division twice) to get

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

Check by expanding: $x^4 - x^3 - 2x^2 + 4x^2 - 4x - 8 = x^4 - x^3 + 2x^2 - 4x - 8$ . ✓

From $x^2 + 4 = 0$ , $x = \pm 2i$ . The four zeros are $2$ , $-1$ , $2i$ and $-2i$ , and

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

The graph shows only the two real zeros. The factor $x^2 + 4$ is always positive, so it never creates an $x$ -intercept.

f(x) = x⁴ − x³ + 2x² − 4x − 8 crosses the x-axis only at x = −1 and x = 2. Its zeros ±2i don't appear on the graph.Open in grapher →

Common mistake

Don't stop when you run out of $x$ -intercepts. A degree- $4$ polynomial always has $4$ complex zeros. If you've found only two real ones, the other two are either another real pair or a complex conjugate pair, and the depressed quadratic will tell you which.

Tip

Use every shortcut the theorems give you. The sum of the coefficients is $f(1)$ , so if they add to $0$ , then $1$ is a zero, and if not, $1$ is out. If all coefficients are positive, Descartes' rule says there are no positive zeros, so test only negative candidates.

Practice

Practice 1

Which number is not a possible rational zero of $f(x) = 3x^3 - 5x^2 + x - 2$ ?

Practice 2

Find all zeros of $f(x) = x^3 - 7x - 6$ .

Separate answers with commas, e.g. 2, -5

Practice 3

Find all zeros of $f(x) = 2x^3 + 3x^2 - 8x + 3$ .

Separate answers with commas, e.g. 2, -5

Practice 4

Find all zeros of $f(x) = x^3 - 5x^2 + 5x + 3$ . Use sqrt() for square roots.

Separate answers with commas, e.g. 2, -5

Practice 5

What does Descartes' rule of signs say about $f(x) = x^5 - 3x^4 + 2x^2 + x - 7$ ?

Practice 6

The polynomial $f(x) = x^3 + x^2 + 3x - 5$ has one real zero and two non-real zeros $a \pm bi$ with $b > 0$ . Enter $(a, b)$ .

Enter a point like (2, -3)

Practice 7

Given that $2i$ is a zero of $f(x) = x^4 + x^3 + 2x^2 + 4x - 8$ , find all the real zeros of $f$ .

Separate answers with commas, e.g. 2, -5

Practice 8

How many non-real complex zeros does $f(x) = x^5 + 3x^3 - 4x$ have?

Enter a number. Fractions like 3/4 and sqrt(2) are OK.