Wize High School Grade 12 Pre-Calculus Textbook > Polynomial Functions

Factoring Polynomials (Degree 3+)

Remainder Theorem, Factor Theorem, & Rational Root Theorem

Factoring polynomials of degree 3 or higher requires the use of polynomial division along with 3 important theorems:

Remainder Theorem

Let $y=f(x)~$ be a polynomial function of degree $n$ . If $f(x)~\div~(x-a)$ , then the remainder, $r$ , is:

\boxed{f(a)=r}

Example 1

Find the remainder of $x^3-5x+1$ when it is divided by $(x-2)$ .

Using the remainder theorem, find $f(2):$

f(2)=(2)^3-5(2)+1\newline{} =-1~~~~~~~~~~~~~

The remainder is -1

Factor Theorem

Let $y=f(x)~$ be a polynomial function of degree $n$ . If $(x-a)$ is a factor of $y=f(x).$ Then:

\boxed{f(a)=0}

Example 2

Determine if $(x+2)$ is a factor of $x^2+5x+6$ .

Using the factor theorem:

f(-2)=(-2)^2+5(-2)+6\newline{} =0~~~~~~~~~~~~~~~~~~~

Since the remainder is 0, then $(x+2)$ is a factor of $x^2+5x+6$

Rational Root Theorem

If $f(x)=ax^n+bx^{n-1}+~...~+cx+d$ is divided by $(x-a)$ , then the possible roots of $f(x)~$ are:

\Bigg\{\frac{\text{Factors of d}}{\text{Factors of a}} \Bigg\}

Example 3

Determine all the roots for $y=2x^3+x-5$ .

Using the rational root theorem:

\Bigg\{\frac{\text{Factors of -5}}{\text{Factors of 2}}\Bigg\}=\pm\displaystyle\frac{1,5}{1,2}=\pm\frac{1}{2},\frac{1}{5},1,5

Factoring Polynomials (Degree 3+)

Use these 3 steps to factor a polynomial:

Use the Rational Root Theorem to determine all possible values of the roots
Use the Remainder & Factor Theorem to test the possible factors for $f(x)$ (Hint: You only need to find one)
Use polynomial division (long or synthetic) to find the other factors of $f(x)$

Example

Fully factor $f(x)=2x^3-2x^2-32x+32$

Step 1.

Use the Rational Root Theorem to determine all possible values of the roots

Factors of 32: 1, 2, 4, 8, 16, 32

Factors of 2: 1, 2

All possible factors are:

\pm\frac{1, 2, 4, 8, 16, 32}{1, 2}=\pm\frac{1}{2},1, 2, 4, 8, 16, 32

Step 2.

Use the Remainder & Factor Theorem to test the possible factors for $f(x)$ (Hint: You only need to find one)

$\underline{x=\frac{1}{2}}$ :

f\bigg(\frac{1}{2}\bigg)=2\bigg(\frac{1}{2}\bigg)^3-2\bigg(\frac{1}{2}\bigg)^2-32\bigg(\frac{1}{2}\bigg)+32\newline{}\newline{} =-\frac{1}{4}-16+32~~~~~~~~~~~~~~~~~~~~~\newline{}\newline{} \neq0~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Since $x=\frac{1}{2}~$ does not give a remainder of 0, then $(2x-1)~$ is not a factor of f(x)

$\underline{x=1}$ :

f(1)=2(1)^3-2(1)^2-32(1)+32\newline{} =0~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Since $x=1~$ gives a remainder of 0, then $(x-1)$ is a factor

Step 3.

Use polynomial division (long or synthetic) to find the other factors of $f(x)$

(2x^3-2x^2-32x+32)~\div~(x-1)

\begin{array}{r| c c c c} &2&-2&-32&32\\ 1&\downarrow&2&0&32\\ \hline &2&0&-32&0 \end{array}

The quotient becomes:

2x^2-32=2(x^2-16)\newline{} ~~~~~~~~~~~~~~~~~~~~~~~~=2(x-4)(x+4)

The division statement becomes:

(2x^3-2x^2-32x+32)=(x+1)2(x-4)(x+4)

Example: Factoring Polynomials (Degree 3+)

Factor fully and graph $x^3-3x^2-2x+6$

Step 1.

Use the Rational Root Theorem to determine all possible values of the roots.

Factors of 6: 1, 2, 3, 6

Factors of 1: 1

All possible roots are:

\pm\frac{1, 2, 4, 8, 16, 32}{1}=\pm1, 2, 3, 6

Step 2.

Use the Remainder & Factor Theorem to test the possible factors for $f(x)$ . (Hint: You only need to find one)

$\underline{x=1}$ :

\begin{array}{rcl} (1)^3-3(1)^2-2(1)+6&=&1-3-2+6\\ &=&2 \end{array}

x = 1 is not a possible root

$\underline{x=3}$ :

\begin{array}{rcl} (3)^3-3(3)^2-2(3)+6&=&27-27-6+6\\ &=&0 \end{array}

Therefore, $x=3$ is a root and $(x-3)$ is a factor

Step 3.

Use polynomial division (long or synthetic) to find the other factors of $f(x).$

(x^3-3x^2-2x+6)~\div~(x-3)

\begin{array}{r| c c c c} &1&-3&-2&6\\ 3&\downarrow&3&0&-6\\ \hline &1&0&-2&0 \end{array}

The quotient becomes:

\begin{array}{rcl} x^2-2&=&(x-\sqrt{2})(x+\sqrt{2})\\ \end{array}

The division statement becomes:

\begin{array}{rcl} x^3-3x^2-2x+6&=&(x-3)(x-\sqrt{2})(x+\sqrt{2})\\ \end{array}

The graph:

Practice: Factoring Polynomials (Degree 3+)

True or False: A factor of $2x^4+12x^3-x^2+8x-3$ is $(x-3)$

Your answer

I don't know

Practice: Factoring Polynomials (Degree 3+)

Fully factor $x^4+x^3-13x^2-25x-12$ .

Your answer

I don't know

Practice: Factoring Polynomials (Degree 3+)

The remainder when $x^3+ax^2+bx+1~$ is divided by (x - 5) is -14. When $x^3+ax^2+bx+1$ is divided by (x+1), the remainder is -2.

What are the values of a and b?

I don't know