« Division of Polynomials 

A polynomial of degree n has at most n distinct zeros.
Let p(x) be a polynomial function with real coefficients. If a + ib is an imaginary zero of p(x), the conjugate abi is also a zero of p(x).
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.
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 posible 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: ${x_1} = 2$, ${x_2} = \frac{1}{6}$, ${x_3} =  5$