8112019 A 26 Remainder

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983209Mathematics Support CentreCoventry University 2001

MATHEMATICS

SUPPORT CENTRE

Title Remainder Theorem and Factor Theorem

Target On completion of this worksheet you should be able to use the remainder

and factor theorems to find factors of polynomials

A26

Generally when a polynomial is divided by a linear

expression there is a remainder

5)123)(2()3543(5

2

3

42

52

63

123

35432

)2()3543(eg

223

2

2

23

2

23

23

+minusminus+=+minus+

minusminus

+minus

minusminus

minusminus

+

minusminus

+minus++

+divide+minus+

x x x x x x

x

x

x x

x x

x x

x x

x x x x

x x x x

Any polynomial can be written in the following

form

polynomial equiv divisor times quotient + remainder

In particular if the divisor is ( x ndash a) and the

polynomial is f ( x) then

f ( x) equiv ( x ndash a) times quotient + remainder

If x = a then

f (a) = (a ndash a) times quotient + remainder

f (a) = remainder

This gives an easy way of finding the remainder

when a polynomial is divided by ( x ndash a)

Examples

1 Using previous example

5)123)(2(

3543)(

2

23

+minusminus+=

+minus+=

x x x

x x x x f

Now using x = -2

5

5)1)2(2)2(3(0

5)1)2(2)2(3)(22()2(

2

2

=

+minusminustimesminusminustimestimes=

+minusminustimesminusminustimes+minus=minus f

ie the remainder

2 Find the remainder when

5311512)1(

0)1(requirewesince 1Substitute

352)(Let

)1( bydividedis )352(

23

23

23

minus=minus+timesminustimes=

=minus=

minus+minus=

minusminus+minus

f

x x

x x x x f

x x x x

The remainder is ndash5

3 Find the remainder when

1017)3(3)3(4)3()3(0)3(requirewesince 3 Substitute

734)(Let

)3( bydividedis )734(

24

24

24

=minusminustimes+minustimes+minus=minus

=+minus=

minus++=

+minus++

f x x

x x x x f

x x x x

The remainder is 101

divisor quotient remainder

Exercise

Find the remainders for the following

)4()4(5

)3()562(4

)1()132(3

)1()234(2

)2()465(1

23

3

234

23

23

minusdivideminusminus

+divideminusminus

+divideminus+minus

minusdivide+++

minusdivideminus+minus

x x x x

x x x

x x x x

x x x x

x x x x

(Answers -4 10 5 -29 -4)

8112019 A 26 Remainder

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983209Mathematics Support CentreCoventry University 2001

Example

Find the remainder when

02151413)1(

1 and 2543)(Let

)1( bydividedis )2543(

23

23

23

=minustimesminustimes+times=

=minusminus+=

minusminusminus+

f

x x x x x f

x x x x

The remainder is 0

quotient)1(

0quotient)1(2543 23

timesminus=

+timesminus=minusminus+

x

x x x x

so )1( minus x is a factor of )2543( 23minusminus+ x x x

We can use the remainder theorem to check for

factors of a polynomial

As before

remainder quotient)()( +timesminus= a x x f

and remainder )( =a f

If )( a x minus is a factor then the remainder is 0

ie 0)( =a f

This is called the factor theorem

We can use the factor theorem to factorise

polynomials although some trial and error is

involved

Examples

1 Is )3( minus x a factor of )3832( 23minusminusminus x x x

Let 3 and )3832()( 23=minusminusminus= x x x x x f

as we are checking whether )3( minus x is a factor

03383332)3(23

=minustimesminustimesminustimes= f so )3( minus x is a factor of )3832( 23

minusminusminus x x x

2 Is )1( minus x a factor of )3832( 23minusminusminus x x x

Using f ( x) as above and x = 1

0123181312)1( 23neminus=minustimesminustimesminustimes= f

so 1minus x is not a factor of 3832 23minusminusminus x x

Exercise

)6103()(

of factora)1( Is5

)6103()(

of factora)3( Is4

)10134()(of factora)2( Is3

)10134()(of factora)2( Is2

)122()(

of factora)1( Is1

23

23

2

2

23

minus++=

minus

minus++=

+

++=minus

++=+

minusminus+=

minus

x x x x f

x

x x x x f

x

x x x f x

x x x f x

x x x x f

x

(Answers yes yes no yes no)

Example

Factorise )652( 23minusminus+ x x x

Let 652)( 23minusminus+= x x x x f Since the

constant is ndash6 we will consider factors of this

ie plusmn 1 plusmn 2 plusmn 3 plusmn 6 We will try )1( minus x

0611512)1( 23=minusminustimes+times= f

so )1( minus x is a factor

Now we can find the quadratic factor by division

or by repeating the above

)32)(2)(1()672)(1(

652)(

0

66

66

77

7

22

672

652)1(

2

23

2

2

23

2

23

++minus=

++minus=

minusminus+=

minus

minus

minus

minus

minus

++

minusminus+minus

x x x x x x

x x x x f

x

x

x x

x x

x x

x x

x x x x

The quadratic factor is factorised in the normal

way

Exercise

Factorise the following

617154)(5

263)(4

433)(3

122)(2

652)(1

23

23

23

23

23

minus+minus=

+++=

minusminusminus=

minusminus+=

minusminus+=

x x x x f

x x x x f

x x x x f

x x x x f

x x x x f

Answers

)34)(2)(1(5

)13)(2(4

)1)(4(3

)12)(1)(1(2

)3)(2)(1(1

2

2

minusminusminus

++

++minus

+minus+

+minus+

x x x

x x

x x x

x x x

8112019 A 26 Remainder

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983209Mathematics Support CentreCoventry University 2001

Example

Find the remainder when

02151413)1(

1 and 2543)(Let

)1( bydividedis )2543(

23

23

23

=minustimesminustimes+times=

=minusminus+=

minusminusminus+

f

x x x x x f

x x x x

The remainder is 0

quotient)1(

0quotient)1(2543 23

timesminus=

+timesminus=minusminus+

x

x x x x

so )1( minus x is a factor of )2543( 23minusminus+ x x x

We can use the remainder theorem to check for

factors of a polynomial

As before

remainder quotient)()( +timesminus= a x x f

and remainder )( =a f

If )( a x minus is a factor then the remainder is 0

ie 0)( =a f

This is called the factor theorem

We can use the factor theorem to factorise

polynomials although some trial and error is

involved

Examples

1 Is )3( minus x a factor of )3832( 23minusminusminus x x x

Let 3 and )3832()( 23=minusminusminus= x x x x x f

as we are checking whether )3( minus x is a factor

03383332)3(23

=minustimesminustimesminustimes= f so )3( minus x is a factor of )3832( 23

minusminusminus x x x

2 Is )1( minus x a factor of )3832( 23minusminusminus x x x

Using f ( x) as above and x = 1

0123181312)1( 23neminus=minustimesminustimesminustimes= f

so 1minus x is not a factor of 3832 23minusminusminus x x

Exercise

)6103()(

of factora)1( Is5

)6103()(

of factora)3( Is4

)10134()(of factora)2( Is3

)10134()(of factora)2( Is2

)122()(

of factora)1( Is1

23

23

2

2

23

minus++=

minus

minus++=

+

++=minus

++=+

minusminus+=

minus

x x x x f

x

x x x x f

x

x x x f x

x x x f x

x x x x f

x

(Answers yes yes no yes no)

Example

Factorise )652( 23minusminus+ x x x

Let 652)( 23minusminus+= x x x x f Since the

constant is ndash6 we will consider factors of this

ie plusmn 1 plusmn 2 plusmn 3 plusmn 6 We will try )1( minus x

0611512)1( 23=minusminustimes+times= f

so )1( minus x is a factor

Now we can find the quadratic factor by division

or by repeating the above

)32)(2)(1()672)(1(

652)(

0

66

66

77

7

22

672

652)1(

2

23

2

2

23

2

23

++minus=

++minus=

minusminus+=

minus

minus

minus

minus

minus

++

minusminus+minus

x x x x x x

x x x x f

x

x

x x

x x

x x

x x

x x x x

The quadratic factor is factorised in the normal

way

Exercise

Factorise the following

617154)(5

263)(4

433)(3

122)(2

652)(1

23

23

23

23

23

minus+minus=

+++=

minusminusminus=

minusminus+=

minusminus+=

x x x x f

x x x x f

x x x x f

x x x x f

x x x x f

Answers

)34)(2)(1(5

)13)(2(4

)1)(4(3

)12)(1)(1(2

)3)(2)(1(1

2

2

minusminusminus

++

++minus

+minus+

+minus+

x x x

x x

x x x

x x x