« Division of Polynomials

Number of Zeros Theorem

A polynomial of degree n has at most n distinct zeros.

Conjugate Zeros Theorem

Let p(x) be a polynomial function with real coefficients. If a + ib is an imaginary zero of p(x), the conjugate a-bi is also a zero of p(x).

The Factor Theorem

For a polynomial f(x) and a constant c,

a. If f(c) = 0, then x - c is a factor of f(x).

b. If x - c is a factor of f(x), then f(c) = 0.

The Factor Theorem tells us that if we find a value of c such that f(c) = 0, then x - c is a factor of f(x). And, if x - c is a factor of f(x), then f(c) = 0.

Rational Root Test

If a polynomial function has integer coefficients, then every rational zero will have the form p/q where p is a factor of the constant and q is a factor of the leading coefficient.

Example 1

Use the Rational Root Test to list all the possible rational zeros for $f(x) = 4{x^3} - 2{x^2} + x + 10$.

Solution:

Step 1: Find factors of the leading coefficient

1, -1, 2, -2, 4, -4

Step 2: Find factors of the constant

1, -1, 2, -2, 5, -5, 10, -10

Step 3: Find all the POSSIBLE rational zeros or roots.

Writing the possible factors as $\frac{p}{q}$ we get:

$\frac{1}{1}$, $\frac{-1}{1}$, $\frac{2}{1}$, $\frac{-2}{1}$, $\frac{4}{1}$, $\frac{-4}{1}$, $\frac{1}{2}$, $\frac{-1}{2}$, $\frac{2}{2}$, $\frac{-2}{2}$, $\frac{4}{2}$, $\frac{-4}{2}$

$\frac{1}{5}$, $\frac{-1}{5}$, $\frac{2}{5}$, $\frac{-2}{5}$, $\frac{4}{5}$, $\frac{-4}{5}$, $\frac{1}{10}$, $\frac{-1}{10}$, $\frac{2}{10}$, $\frac{-2}{10}$, $\frac{4}{10}$, $\frac{-4}{10}$

Here is a final list of all the possible rational zeros, each one written once and reduced:

$1$, $-1$, $2$, $-2$, $4$, $-4$, $\frac{1}{2}$, $\frac{-1}{2}$, $\frac{1}{5}$, $\frac{-1}{5}$, $\frac{4}{5}$, $\frac{-4}{5}$, $\frac{1}{10}$, $\frac{-1}{10}$

Example 2

Factor f(x) = $f(x) = 6{x^3} + 17{x^2} - 63x + 10$into linear factors

Solution

Step 1: Find factors of the leading coefficient

1, -1, 2, -2, 3, -3, 6, -6

Step 2: Find factor of the constant

1, -1, 2, -2, 5, -5, 10, -10

Step 3: Find all the possible rational zeros or roots.

Writing the possible factors as $\frac{p}{q}$ we get:

$\frac{1}{1}$, $\frac{-1}{1}$, $\frac{2}{1}$, $\frac{-2}{1}$, $\frac{3}{1}$, $\frac{-3}{1}$, $\frac{6}{1}$, $\frac{-6}{1}$, $\frac{1}{2}$, $\frac{-1}{2}$, $\frac{2}{2}$, $\frac{-2}{2}$, $\frac{3}{2}$, $\frac{-3}{2}$, $\frac{6}{2}$, $\frac{-6}{2}$

$\frac{1}{5}$, $\frac{-1}{5}$, $\frac{2}{5}$, $\frac{-2}{5}$, $\frac{3}{5}$, $\frac{-3}{5}$, $\frac{6}{5}$, $\frac{-6}{5}$, $\frac{1}{10}$, $\frac{-1}{10}$, $\frac{2}{10}$, $\frac{-2}{10}$, $\frac{3}{10}$, $\frac{-3}{10}$, $\frac{6}{10}$, $\frac{-6}{10}$

We check that -5 is the zero of $f(x)$.

Now we use the Factor Theorem

$$\frac{{6{x^3} + 17{x^2} - 63x + 10}}{{x + 5}} = 6{x^2} - 13x + 2$$

Now we have to solve $6x^2 - 13x + 2 = 0.$

${x_{1,2}} = \frac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} = \frac{{13 \pm \sqrt {{{( - 13)}^2} - 4 \cdot 6 \cdot 2} }}{{2 \cdot 6}}$

${x_1} = 2$, ${x_2} = \frac{1}{6}$

The roots are: ${x_1} = 2$, ${x_2} = \frac{1}{6}$, ${x_3} = - 5$