# 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) =a_{0}x+^{n}a_{1}x^{n}^{−1}+a_{2}x^{n}^{−2}+ ... +a_{n}

where

a_{0}≠ 0 and

nis a positive integer, called thedegreeof the polynomial.

**Example 1 **

*f*(*x*) = 7*x*^{5} + 4*x*^{3} − 2*x*^{2} − 8*x* + 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`

Another way of thinking about this example is:

`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*) = 3*x*^{2} + 5*x *− 8 by (*x* − 2).

Answer

`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:

3

x^{2}+ 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 (x−r) and a remainderRis obtained, thenf(r) =R.

### Example 3

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

Answer

Since we are dividing *f*(*x*) = 3*x*^{2} + 5*x *− 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

3

x^{3}−x^{2}− 20x+ 5

is divided by (*x* + 4).

Answer

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`

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## The Factor Theorem

The Factor Theorem states:

If the remainder

f(r) =R= 0, then (x−r) is a factor off(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*) =* x*^{3} + 2*x*^{2} − 5*x *− 6?

Answer

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)`.

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### Exercises

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

(2

x^{4}− 10x^{2}+ 30x- 60) ÷ (x+ 4)

Answer

`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*) = 2*x*^{4} − 10*x*^{2} + 30*x* − 60

Remainder = *f*(−4) = 2(-4)^{4} − 10(−4)^{2} + 30(−4) − 60 = 172

This is the same answer we achieved by long division.

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**2.** Find the remainder using the Remainder Theorem

(

x^{4}− 5x^{3}+x^{2}− 2x+ 6) ÷ (x+ 4)

Answer

Applying the Remainder Theorem:

*f*(*x*) = *x*^{4} − 5*x*^{3} + *x*^{2} − 2*x* + 6

*f*(−4) = (−4)^{4} − 5(−4)^{3} + (−4)^{2} − 2(−4) + 6 = 606

So the remainder is `606`.

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

f(x) =x^{5}− 2x^{4}+ 3x^{3}− 6x^{2}− 4x+ 8.

Answer

*f*(*x*) = *x*^{5} − 2*x*^{4} + 3*x*^{3} − 6*x*^{2} − 4*x* + 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.

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

f(x) = 2x^{3}+ 3x^{2}− 8x− 12.

Answer

`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 2*x*^{3} + 3*x*^{2} − 8*x* − 12, since the function value is `0`.

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