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`

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) = 3x2 + 5x − 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:

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).

Answer

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.

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Example 4

By using the remainder theorem, determine the remainder when

3x3x2 − 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`

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?

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)`.

Exercises

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

(2x4 − 10x2 + 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) = 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)

Answer

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.

Answer

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.

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

f(x) = 2x3 + 3x2 − 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 2x3 + 3x2 − 8x − 12, since the function value is `0`.