# 2. The Remainder Theorem and the Factor Theorem

This section discusses the historical method of solving higher degree polynomial equations.

As we discussed in the previous section Polynomial Functions and Equations, a polynomial function is of the form:

f(x) = a0xn + a1xn−1 + a2xn−2 + ... + an

where

a0 ≠ 0 and

n is a positive integer, called the degree of the polynomial.

### Example 1

f(x) = 7x5 + 4x3 − 2x2 − 8x + 1 is a polynomial function of degree 5.

## Dividing Polynomials

First, let's consider what happens when we divide numbers.

Example (a): Say we try to divide 13 by 5. We will get the answer 2 and have a remainder of 3. We could write this as:

13/5 = 2 + 3/5

13 = 2 × 5 + 3

Example (b), Long Division: In primary school, you may have learned to divide larger numbers as follows. Let's divide 3,756 by 23.

 163 23 {:) 3756 23 We multiply 23 by 1 = 23. 145 37-23 = 14. Then bring down the 5. 138 Multiply 23 by 6=138. 76 145-138=7. Bring down the 6. 69 Multiply 23 by 3=69. 7 76-69=7. This is the remainder.

So we can conclude 3,756 -: 23 = 163 + 7/23, or putting it another way, 3,756 = 163xx23 + 7.

Division of polynomials is an extension of our number examples.

If we divide a polynomial by (x r), we obtain a result of the form:

f(x) = (x r) q(x) + R

where q(x) is the quotient and R is the remainder.

Let's now see an example of polynomial division.

### Example 2

Divide f(x) = 3x2 + 5x − 8 by (x − 2).

 3x+11 x-2 {:) 3x^2+5x-8 3x^2-6x We multiply (x-2) by 3x =  3x^2-6x, giving 3x^2 as the first term. 11x-8 5x-(-6x)  = 5x+6x =11x. Then bring down the -8. 11x-22 Multiply (x-2) by 11= 11x-22. 14 -8-(-22)  = 14. This is the remainder.

Thus, we can conclude that:

3x2 + 5x − 8 = (x − 2)(3x + 11) + 14

where the quotient q(x) = 3x + 11 and the remainder R = 14.

We can also write our answer as:

(3x^2+5x-8)-:(x-2) =(3x+11)+14/(x-2

## The Remainder Theorem

Consider f(x) = (x r)q(x) + R

Note that if we let x = r, the expression becomes

f(r) = (r r) q(r) + R

Simplifying gives:

f(r) = R

This leads us to the Remainder Theorem which states:

If a polynomial f(x) is divided by (xr) and a remainder R is obtained, then f(r) = R.

### Example 3

Use the remainder theorem to find the remainder for Example 1 above, which was divide f(x) = 3x2 + 5x − 8 by (x − 2).

Since we are dividing f(x) = 3x2 + 5x − 8 by (x − 2), we let

x = 2.

Hence, the remainder, R is given by:

R=f(2)=3(2)^2+5(2)-8=14

This is the same remainder we achieved before.

### Example 4

By using the remainder theorem, determine the remainder when

3x3x2 − 20x + 5

is divided by (x + 4).

If we divide by (x − r), we let x = r.

Hence, since we are dividing by (x + 4), we let x = -4.

Therefore the remainder

R=f(-4)

=3(-4)^3-(-4)^2 -20(-4)+5

=-192-16+80+5

=-123

## The Factor Theorem

The Factor Theorem states:

If the remainder f(r) = R = 0, then (x r) is a factor of f(x).

The Factor Theorem is powerful because it can be used to find roots of polynomial equations.

### Example 5

Is (x + 1) a factor of f(x) = x3 + 2x2 − 5x − 6?

In this case we need to test the remainder r = -1.

R= f(r)

= f(-1)

= (-1)^3+ 2(-1)^2- 5(-1) - 6

= -1 + 2 + 5 - 6

= 0

Therefore, since f(-1) = 0, we conclude that (x + 1) is a factor of f(x).

### Exercises

1. Find the remainder R by long division and by the Remainder Theorem.

(2x4 − 10x2 + 30x - 60) ÷ (x + 4)

 2x^3-8x^2+22x-58 x+4 {:) 2x^4+0x^3-10x^2+30x-60 2x^4+8x^3 -8x^3-10x^2 -8x^3-32x^2 22x^2+30x 22x^2+88x -58x-60 -58x-232 172

From the above working, we conclude the remainder is 172.

Now, using the Remainder Theorem:

f(x) = 2x4 − 10x2 + 30x − 60

Remainder = f(−4) = 2(-4)4 − 10(−4)2 + 30(−4) − 60 = 172

This is the same answer we achieved by long division.

2. Find the remainder using the Remainder Theorem

(x4 − 5x3 + x2 − 2x + 6) ÷ (x + 4)

Applying the Remainder Theorem:

f(x) = x4 − 5x3 + x2 − 2x + 6

f(−4) = (−4)4 − 5(−4)3 + (−4)2 − 2(−4) + 6 = 606

So the remainder is 606.

3. Use the Factor Theorem to decide if (x − 2) is a factor of

f(x) = x5 − 2x4 + 3x3 − 6x2 − 4x + 8.

f(x) = x5 − 2x4 + 3x3 − 6x2 − 4x + 8

f(2) = (2)5 − 2(2)4 + 3(2)3 6(2)2 4(2) + 8 = 0

Since f(2) = 0, we can conclude that (x - 2) is a factor.

4. Determine whether -3/2 is a zero (root) of the function:

f(x) = 2x3 + 3x2 − 8x − 12.

f(-3/2)=2(-3/2)^3+3(-3/2)^2-8(-3/2) -12

=(-27/4)+(27/4)+12-12

= 0

So yes, -3/2 is a root of 2x3 + 3x2 − 8x − 12, since the function value is 0.

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