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

Factoring Polynomials (Degree 3+)

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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) y=f(x)~y=f(x)  be a polynomial function of degree nnn. If f(x) ÷ (x−a)f(x)~\div~(x-a)f(x) ÷ (x−a), then the remainder, rrr, is: 
f(a)=r\boxed{f(a)=r}f(a)=r​

Example 1

Find the remainder of x3−5x+1x^3-5x+1x3−5x+1 when it is divided by (x−2)(x-2)(x−2). 

Using the remainder theorem, find f(2):f(2):f(2):
f(2)=(2)3−5(2)+1=−1             f(2)=(2)^3-5(2)+1\newline{}
=-1~~~~~~~~~~~~~f(2)=(2)3−5(2)+1=−1             
The remainder is -1

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Factor Theorem
Let y=f(x) y=f(x)~y=f(x)  be a polynomial function of degree nnn. If (x−a)(x-a)(x−a)is a factor of y=f(x).y=f(x).y=f(x).Then:
f(a)=0\boxed{f(a)=0}f(a)=0​
 
Example 2

Determine if (x+2)(x+2)(x+2) is a factor of x2+5x+6x^2+5x+6x2+5x+6 .

Using the factor theorem:
f(−2)=(−2)2+5(−2)+6=0                   f(-2)=(-2)^2+5(-2)+6\newline{}
=0~~~~~~~~~~~~~~~~~~~f(−2)=(−2)2+5(−2)+6=0                   

Since the remainder is 0, then (x+2)(x+2)(x+2) is a factor of x2+5x+6x^2+5x+6x2+5x+6

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Rational Root Theorem
If f(x)=axn+bxn−1+ ... +cx+df(x)=ax^n+bx^{n-1}+~...~+cx+df(x)=axn+bxn−1+ ... +cx+d is divided by (x−a)(x-a)(x−a), then the possible roots of f(x) f(x)~f(x)  are:
{Factors of dFactors of a}\Bigg\{\frac{\text{Factors of d}}{\text{Factors of a}}

\Bigg\}{Factors of aFactors of d​}

Example 3

Determine all the roots for y=2x3+x−5y=2x^3+x-5y=2x3+x−5.

Using the rational root theorem:
{Factors of -5Factors of 2}=±1,51,2=±12,15,1,5\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{Factors of 2Factors of -5​}=±1,21,5​=±21​,51​,1,5

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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)f(x)f(x) (Hint: You only need to find one)
Use polynomial division (long or synthetic) to find the other factors of f(x)f(x)f(x)
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Example
Fully factor f(x)=2x3−2x2−32x+32f(x)=2x^3-2x^2-32x+32f(x)=2x3−2x2−32x+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:
±1,2,4,8,16,321,2=±12,1,2,4,8,16,32\pm\frac{1, 2, 4, 8, 16, 32}{1, 2}=\pm\frac{1}{2},1, 2, 4, 8, 16, 32±1,21,2,4,8,16,32​=±21​,1,2,4,8,16,32

Step 2.
Use the Remainder & Factor Theorem to test the possible factors for f(x)f(x)f(x) (Hint: You only need to find one)

x=12‾\underline{x=\frac{1}{2}}x=21​​:
f(12)=2(12)3−2(12)2−32(12)+32=−14−16+32                     ≠0                                           
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~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

f(21​)=2(21​)3−2(21​)2−32(21​)+32=−41​−16+32                     =0                                           

Since x=12 x=\frac{1}{2}~x=21​  does not give a remainder of 0, then (2x−1) (2x-1)~(2x−1)  is not a factor of f(x)


x=1‾\underline{x=1}x=1​:
f(1)=2(1)3−2(1)2−32(1)+32=0                                   f(1)=2(1)^3-2(1)^2-32(1)+32\newline{}
=0~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~f(1)=2(1)3−2(1)2−32(1)+32=0                                   

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

Step 3.
Use polynomial division (long or synthetic) to find the other factors of f(x)f(x)f(x)
(2x3−2x2−32x+32) ÷ (x−1)(2x^3-2x^2-32x+32)~\div~(x-1)(2x3−2x2−32x+32) ÷ (x−1)

2−2−32321↓203220−320\begin{array}{r| c c c c}
&2&-2&-32&32\\
1&\downarrow&2&0&32\\
\hline
&2&0&-32&0
\end{array}1​2↓2​−220​−320−32​32320​​
The quotient becomes:
2x2−32=2(x2−16)                        =2(x−4)(x+4)2x^2-32=2(x^2-16)\newline{}
~~~~~~~~~~~~~~~~~~~~~~~~=2(x-4)(x+4)2x2−32=2(x2−16)                        =2(x−4)(x+4)

The division statement becomes:
(2x3−2x2−32x+32)=(x+1)2(x−4)(x+4)(2x^3-2x^2-32x+32)=(x+1)2(x-4)(x+4)(2x3−2x2−32x+32)=(x+1)2(x−4)(x+4)

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Example: Factoring Polynomials (Degree 3+)

Factor fully and graph x3−3x2−2x+6x^3-3x^2-2x+6x3−3x2−2x+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:
±1,2,4,8,16,321=±1,2,3,6\pm\frac{1, 2, 4, 8, 16, 32}{1}=\pm1, 2, 3, 6±11,2,4,8,16,32​=±1,2,3,6


Step 2.
Use the Remainder & Factor Theorem to test the possible factors for f(x)f(x)f(x). (Hint: You only need to find one)

x=1‾\underline{x=1}x=1​:
(1)3−3(1)2−2(1)+6=1−3−2+6=2\begin{array}{rcl}
(1)^3-3(1)^2-2(1)+6&=&1-3-2+6\\
&=&2

\end{array}(1)3−3(1)2−2(1)+6​==​1−3−2+62​
x = 1 is not a possible root


x=3‾\underline{x=3}x=3​:
(3)3−3(3)2−2(3)+6=27−27−6+6=0\begin{array}{rcl}
(3)^3-3(3)^2-2(3)+6&=&27-27-6+6\\
&=&0

\end{array}(3)3−3(3)2−2(3)+6​==​27−27−6+60​
Therefore, x=3x=3x=3 is a root and (x−3)(x-3)(x−3) is a factor


Step 3.
Use polynomial division (long or synthetic) to find the other factors of f(x).f(x).f(x).

(x3−3x2−2x+6) ÷ (x−3)(x^3-3x^2-2x+6)~\div~(x-3)(x3−3x2−2x+6) ÷ (x−3)

1−3−263↓30−610−20\begin{array}{r| c c c c}
&1&-3&-2&6\\
3&\downarrow&3&0&-6\\
\hline
&1&0&-2&0
\end{array}3​1↓1​−330​−20−2​6−60​​
The quotient becomes:
x2−2=(x−2)(x+2)\begin{array}{rcl}
x^2-2&=&(x-\sqrt{2})(x+\sqrt{2})\\
\end{array}x2−2​=​(x−2​)(x+2​)​

The division statement becomes:
x3−3x2−2x+6=(x−3)(x−2)(x+2)\begin{array}{rcl}
x^3-3x^2-2x+6&=&(x-3)(x-\sqrt{2})(x+\sqrt{2})\\
\end{array}x3−3x2−2x+6​=​(x−3)(x−2​)(x+2​)​
The graph:

Practice: Factoring Polynomials (Degree 3+)
True or False: A factor of 2x4+12x3−x2+8x−32x^4+12x^3-x^2+8x-32x4+12x3−x2+8x−3 is (x−3)(x-3)(x−3)

Answer

I don't know

Practice: Factoring Polynomials (Degree 3+)
Fully factor x4+x3−13x2−25x−12x^4+x^3-13x^2-25x-12x4+x3−13x2−25x−12.

Answer

I don't know

Practice: Factoring Polynomials (Degree 3+)
The remainder when x3+ax2+bx+1 x^3+ax^2+bx+1~x3+ax2+bx+1  is divided by (x - 5) is -14. When x3+ax2+bx+1x^3+ax^2+bx+1x3+ax2+bx+1 is divided by (x+1), the remainder is -2. 

What are the values of a and b?

I don't know

Extra Practice

Polynomial Factoring

Factor the polynomial f(x)=x4+2x3−3x2+4x+12f(x)=x^4+2x^3-3x^2+4x+12f(x)=x4+2x3−3x2+4x+12. 

Factoring Polynomials (Degree 3+)

Practice: Factoring Polynomials (Degree 3+)
The remainder when x3+ax2+bx−2 x^3+ax^2+bx-2~x3+ax2+bx−2  is divided by (x - 2) is 5. When x3+ax2+bx+2x^3+ax^2+bx+2x3+ax2+bx+2 is divided by (x+2), the remainder is 8. 
What are the values of a and b?

Factoring Polynomials (Degree 3+)

Practice: Factoring Polynomials (Degree 3+)
The remainder when x3+ax2+bx−1 x^3+ax^2+bx-1~x3+ax2+bx−1  is divided by (x - 1) is -10. When x3+ax2+bx+3x^3+ax^2+bx+3x3+ax2+bx+3 is divided by (x+1), the remainder is -7. 
What are the values of a and b?

Factoring Polynomials (Degree 3+)

Practice: Factoring Polynomials (Degree 3+)
Fully factor 3x3−2x2−3x+23x^3-2x^2-3x+23x3−2x2−3x+2.

Factoring Polynomials (Degree 3+)

Practice: Factoring Polynomials (Degree 3+)
Fully factor 2x3−15x2+22x+152x^3-15x^2+22x+152x3−15x2+22x+15.

Factoring Polynomials (Degree 3+)

Practice: Factoring Polynomials (Degree 3+)
True or False: A factor of 5x5−3x4+6x3−x2+105x^5-3x^4+6x^3-x^2+105x5−3x4+6x3−x2+10 is (2x−1)(2x-1)(2x−1)

Factoring Polynomials (Degree 3+)

Practice: Factoring Polynomials (Degree 3+)
True or False: A factor of x4+x3−7x2−x+6x^4+x^3-7x^2-x+6x4+x3−7x2−x+6 is (x+3)(x+3)(x+3).