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10&10a

4.08 Polynomial long division

Lesson

Introduction

 Polynomials  can be added, subtracted and multiplied together. They can also be divided.

We're familiar with the idea of  factorising quadratic polynomials  . This idea can be extended to polynomials of any degree. That is, a polynomial of any degree can be rewritten as a product of polynomials of a lesser degree.

Quadratic polynomials, can sometimes be factorised into the product of two linear polynomials. That is, a polynomial of degree 2 can be factorised into two polynomials of degree 1.

For a polynomial of any degree, the sum of the degrees of the factors will equal the degree of the original polynomial. For example, a polynomial of degree 4 can be factorised into 4 polynomials of degree 1 or 2 polynomials of degree 2.

If A(x) is a factor of P(x) then that means that A(x) evenly divides P(x). If P(x)=A(x)Q(x) then it follows that \dfrac{P(x)}{A(x)}=Q(x). This is how we can divide polynomials.

The remainder theorem

Not all polynomials can be evenly divided by other polynomials, in the same way that not all integers can be evenly divided by other integers. However, in the case where they do not evenly divide, there will be a remainder which will also be a polynomial.

In general, for any polynomials P(x) and A(x) there will be polynomials Q(x) and R(x) such that P(x)=A(x)Q(x)+R(x). The sum of the degrees of A(x) and Q(x) will equal the degree of P(x) and the degree of R(x) will be less than the degree of A(x).

If A(x)=x-a for some number a then R(x)=P(a). In other words, the remainder will be the value of P(x) with a substituted for x. This is called the remainder theorem.

Examples

Example 1

Find the remainder when P(x)=-4x^{4}+6x^{3}+4x^{2}-7x+7 is divided by A(x)=3x-1.

Worked Solution
Create a strategy

Equate A(x) to 0 to find the value of x to substitute into P(x).

Apply the idea
\displaystyle 3x-1\displaystyle =\displaystyle 0Equate A(x) to 0
\displaystyle 3x\displaystyle =\displaystyle 1Add 1 to both sides
\displaystyle x\displaystyle =\displaystyle \dfrac{1}{3}Divide both sides by 3
\displaystyle P(x)\displaystyle =\displaystyle -4x^{4}+6x^{3}+4x^{2}-7x+7Write the polynomial
\displaystyle P\left( \dfrac{1}{3}\right)\displaystyle =\displaystyle -4\left(\dfrac{1}{3}\right)^{4}+6\left(\dfrac{1}{3}\right)^{3}+4\left(\dfrac{1}{3}\right)^{2}-7\left(\dfrac{1}{3}\right)+7Substitute x=\dfrac{1}{3}
\displaystyle =\displaystyle \dfrac{428}{81}Evaluate
\displaystyle =\displaystyle 5\dfrac{23}{81}Rewrite as a mixed number

The remainder is 5\dfrac{23}{81}.

Idea summary

For polynomials P(x) and A(x) there are polynomials Q(x) and R(x) such that

\displaystyle P(x)=A(x)Q(x)+R(x)
\bm{A(x)}
is the divisor
\bm{Q(x)}
is the quotient
\bm{R(x)}
is the remainder

The remainder theorem says that if A(x)=x-a then R(x)=P(a).

The factor theorem

If A(x) is a factor of P(x) then the remainder R(x)=0. From the remainder theorem, this means that if A(x)=x-a then P(a)=0. The converse is also true, so if P(a)=0 then x-a is a factor of P(x). This is called the factor theorem. We call values of a where P(a)=0 zeros of the polynomial P(x).

Examples

Example 2

Using the factor theorem or otherwise, rewrite x^{3}-6x^{2}+11x-6 as a product of linear factors.

Worked Solution
Create a strategy

Substitute factors of -6 into the polynomial to find the linear factors.

Apply the idea

Integer factors of -6 are: \pm 1,\,\pm 2,\,\pm 3 and \pm 6.

Let's try the factor 2:

\displaystyle P(x)\displaystyle =\displaystyle x^3-6x^2+11x-6Let P(x) be the polynomial
\displaystyle P(2)\displaystyle =\displaystyle (2)^3-6\times (2)^2+11\times(2)-6Substitute x=2
\displaystyle =\displaystyle 0Evaluate

This means that (x-2) is a factor of x^3-6x^2+11x-6.

Let's try the factor -1:

\displaystyle P(x)\displaystyle =\displaystyle x^3-6x^2+11x-6Let P(x) be the polynomial
\displaystyle P(-1)\displaystyle =\displaystyle (-1)^3-6\times (-1)^2+11\times(-1)-6Substitute x=-1
\displaystyle =\displaystyle -24Evaluate

This means that (x+1) is not a factor of x^3-6x^2+11x-6.

Let's try the factor 1:

\displaystyle P(x)\displaystyle =\displaystyle x^3-6x^2+11x-6Let P(x) be the polynomial
\displaystyle P(1)\displaystyle =\displaystyle (1)^3-6\times (1)^2+11\times(1)-6Substitute x=1
\displaystyle =\displaystyle 0Evaluate

This means that (x-1) is a factor of x^3-6x^2+11x-6.

Let's try the factor 3:

\displaystyle P(x)\displaystyle =\displaystyle x^3-6x^2+11x-6Let P(x) be the polynomial
\displaystyle P(3)\displaystyle =\displaystyle (3)^3-6\times (3)^2+11\times(3)-6Substitute x=3
\displaystyle =\displaystyle 0Evaluate

This means that (x-3) is a factor of x^3-6x^2+11x-6.

So using these factors we get the following factorisation: x^{3}-6x^{2}+11x-6=(x-2)(x-1)(x-3)

Idea summary

The factor theorem says that if P(a)=0, then A(x)=x-a is a factor of P(x).

The converse of the factor theorem says that if A(x)=x-a is a factor of P(x) then P(a)=0.

Polynomial long division

A method we can use to divide polynomials is called polynomial long division. This works similarly to long division of numbers. Watch this video to learn how to do polynomial long division.

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Examples

Example 3

If the distance travelled is (3x^{3}-2x^{2}+4x+9) kilometres and the speed is (x+1)\text{ km/h}, what is the time travelled in hours?

Worked Solution
Create a strategy

By using the long division, divide the distance by the speed to find the time travelled in hours.

Apply the idea

We need to divide (3x^{3}-2x^{2}+4x+9) by x+1.

This image shows the steps of dividing a polynomial by x plus 1 using long division. Ask your teacher for more information.

Set up the long division.

This image shows the steps of dividing a polynomial by x plus 1 using long division. Ask your teacher for more information.

Divide 3x^3 by x to get 3x^2.

This image shows the steps of dividing a polynomial by x plus 1 using long division. Ask your teacher for more information.

Multiply 3x^2 by x+1 to get 3x^3+3x^2. Write the expression under the first two terms of the polynomial.

This image shows the steps of dividing a polynomial by x plus 1 using long division. Ask your teacher for more information.

Subtract 3x^3+3x^2 from 3x^{3}-2x^{2}+4x+9 to get -5x^2+4x+9.

This image shows the steps of dividing a polynomial by x plus 1 using long division. Ask your teacher for more information.

Divide -5x^2 by x to get -5x.

Multiply -5x by x+1 to get -5x^2-5x.

Subtract -5x^2-5x from -5x^2+4x+9 to get 9x+9.

This image shows the steps of dividing a polynomial by x plus 1 using long division. Ask your teacher for more information.

Divide 9x by x to get 9.

Multiply 9 by x+1 to get 9x+9.

Subtract 9x+9 from 9x+9 to get 0 remainder.

The time travelled is given by 3x^2-5x+9 hours.

Idea summary

We can divide polynomials using polynomial long division.

Outcomes

ACMNA266 (10a)

Investigate the concept of a polynomial and apply the factor and remainder theorems to solve problems

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