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Wize Grade 11 Mathematics Textbook > Polynomials Factoring and Graphs

Factoring Polynomials

Factoring Special TrinomialsPrevious Section
Graphs of PolynomialsNext Section

The Remainder Theorem



If we divide a polynomial by a factor such as x−ax- a , the remainder is the same as evaluating the polynomial directly as in f(a)f(a).

f(a)=r\boxed{f(a) = r}


Example 1

Verify the remainder theorem works for the following polynomial and value.
  • f(x)=7x3−18x2−4x−15f(x) = 7x^3 - 18x^2 - 4x - 15
  • a=3a = 3
We can divide using synthetic division. This shows that the remainder is 0.


Evaluating the polynomial directly we have

f(3)=7(3)3−18(3)2−4(3)−15=189−162−12−15=0\begin{aligned} f(3)&=7(3)^3 - 18(3)^2 - 4(3) - 15 \\ &= 189 -162 - 12 - 15 \\ &= 0 \end{aligned}
In both cases we have 0.


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Example 2

When f(x)=x2−x−15f(x) = x^2 - x - 15 , was divided by (x−7)(x - 7) the remainder was 2727. Use this to evaluate

f(7)f(7)

From the remainder theorem we know the remainder will be the same as evaluating the function directly so we have
f(7)=27f(7)=27

The Factor Theorem

The Factor Theorem

The factor theorem says that (x−a)(x - a) is a factor of f(x)f(x) if and only if f(a)=0f(a) = 0.

We can use this and the remainder theorem to test possible factors of a polynomial.

Example 1

Determine if (x+4)(x + 4) is a factor of the following polynomials.

1. f(x)=x4+2x3−5x2+8x−16f(x) = x^4 + 2x^3 - 5x^2 + 8x - 16

Evaluating the function using synthetic division we have that f(−4)=0f(-4) = 0 .


This means (x+4)(x + 4) is a factor. We can write out the factored polynomial:

(x+4)(x3−2x2+3x−4)(x+4)(x^3 - 2x^2 + 3x - 4)


2. g(x)=x4+2x3−5x2+5x−14g(x) = x^4 + 2x^3 - 5x^2 + 5x - 14

Evaluating the function using synthetic division we have that g(−4)=14g(-4) = 14.


Since this is not zero, it means that (x+4)(x+4) is not a factor.

Example: Factoring Polynomials

Factor the following polynomial

g(x)=3x4+11x3−19x2+5xg(x) = 3x^4 + 11x^3 - 19x^2 + 5x

Given that f(1)=0f(1) = 0

First we can see that all of the terms have a common factor of xx, so we want to factor this first.

x(3x3+11x2−19x+5)x(3x^3 + 11x^2 - 19x + 5)

With the remaining polynomial we can use the information that f(1)=0f(1) = 0.
This tell us that (x−1)(x - 1) is a factor.

From synthetic division we have


From the last row we can now write the polynomial as

x(x−1)(3x2+14x−5)x(x - 1)(3x^2 + 14x - 5)

The remaining polynomial is a trinomial. Factoring this we arrive at

x(x−1)(x+5)(3x−1)x(x - 1)(x + 5)(3x - 1)


Practice: Factoring Polynomials

Use synthetic division and the factor theorem to quickly determine if the following are factors of the polynomial

x3+3x2−28x−60x^3 + 3x^2 - 28x - 60
1. (x+6)(x + 6)

2. (x−5)(x - 5)

3. (x−2)(x - 2)

Practice: Factoring Polynomials

The following polynomial
f(x)=16x3−112x2−x+7f(x) = 16x^3 - 112x^2 - x + 7

has a factor of (x−7)(x - 7).
Use this information to find the other two factors.

When you are done, write the entire polynomial in factored form.

Practice: Factoring Polynomials

Use the remainder theorem to determine if (x+1)(x + 1)is a factor of the polynomial
f(x)=x31+x+2f(x) = x^{31} + x +2