22: Division and The Remainder Theorem © Christine Crisp “Teach A Level Maths” Vol. 1: AS Core Modules.

22: Division and22: Division andThe Remainder The Remainder

TheoremTheorem

© Christine Crisp

““Teach A Level Maths”Teach A Level Maths”

Vol. 1: AS Core Vol. 1: AS Core ModulesModules

Division and the Remainder Theorem

Module C1AQA Edexc

elOCR

MEI/OCR

Module C2

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Division and the Remainder Theorem

We’ll first look at what happens when we divide numbers.

21327 e.g.

Quotient

2

13

2

7This can be written

as

Remainder

3 is called the quotient 1 is the remainder

Algebraic expressions can be divided in a similar way

Division

Division and the Remainder Theorem

2

3 1

x

x 22

3 1

xx

x

2

1

xx

Quotient

Remainder

e.g. 1 Find the quotient and remainder when

is divided by

13 x

2x

Solution:

e.g. 2 Write in the form 3

3 532

x

xx 3x

CBxA

Solution: 33

3

3

3 532532

x

x

x

x

x

xx

3

532

x

x

Division and the Remainder TheoremExercis

es Find the quotient and remainder when is divided by x

47 23 xx1.

xx

x

x

x 47 23

xxx

472

x

xx 47 23Solution:

Solution: 2

23 342

x

xxx 22

2

2

3 342

x

x

x

x

x

x

2

342

x

xx

The quotient is and the remainder is -4

xx 72

2. Write in the form 2x

DCxBAx

2

23 342

x

xxx

Division and the Remainder Theorem

x2

Dividing by an expression of the form x - a can be done in 2 ways

Solution: Method 1 Long division.

• Divide the 1st term of the numerator by the 1st term of the denominator.

Write the division as follows:

e.g.1 Divide by62 x 1x

621 xxx 2

Division and the Remainder Theorem

x2

621 xx

Write the division as follows:

• Write this answer above the polynomial being divided.

2

e.g.1 Divide by62 x 1x

Dividing by an expression of the form x - a can be done in 2 ways

Solution: Method 1 Long division.

x 2

Division and the Remainder Theorem

621 xx

Write the division as follows: 2

e.g.1 Divide by62 x 1x

Dividing by an expression of the form x - a can be done in 2 ways

• Multiply x – 1 by this number . . . and write the answer

below• Subtract:6 – (– 2) = 8

8

The quotient is 2 and the remainder is 8.

So,1

82

1

62

xx

x

22 x

)1(2 x

22 x

Solution: Method 1 Long division.

Division and the Remainder Theorem

1

x

)1( x

Dividing by an expression of the form x - a can be done in 2 ways

Solution: Method 2 ( Inspection )

e.g.1 Divide by62 x 1x

1

62

x

x• Write

• Copy the denominator onto the top line

Division and the Remainder Theorem

1

x

Solution: Method 2 ( Inspection )

2 )1( xx2

• Divide the 1st term of the numerator . . .

Dividing by an expression of the form x - a can be done in 2 wayse.g.1 Divide by62 x 1x

• Write

• Multiply . . .

. . . by the 1st term of the denominator

x 21

62

x

x

so the 1st term at the top is now correct2x = 2x

Division and the Remainder Theorem

1

x1

62

x

x 82 )1( x6

Dividing by an expression of the form x - a can be done in 2 wayse.g.1 Divide by62 x 1x

• Write

• Adjust the constant term . . .

2+ 6 =

1

8

1

)1(2

1

62

xx

x

x

x• Separate the 2 terms:

The quotient is 2 and the remainder is 8

+ 8

Solution: Method 2 ( Inspection )

Division and the Remainder Theorem

Method 1 is very complicated for harder divisions, so from now on we

will use Method 2 only.

Division and the Remainder Theorem

)1( x

Solution:

e.g.2 Divide by352 2 xx 1x

1

352 2

x

xx

1x

Write the denominator on the top line

Division and the Remainder Theorem

1

352 2

x

xx )1( x22x

Solution:

e.g.2 Divide by352 2 xx 1x

1x

x2

Correct the 1st term.

22x 22x

Division and the Remainder Theorem

1

352 2

x

xx )1( x )1( x

Solution:

e.g.2 Divide by352 2 xx 1x

1x

x2x5

x5 x2 x7

7

• Copy the denominator and correct the next term.

Division and the Remainder Theorem

)1( x )1( x

Solution:

e.g.2 Divide by352 2 xx 1x

1

352 2

x

xx

1x

x2

• Correct the last term . . . 3 7 4

7 43

• Check the numerator.47722 2 xxx 352 2 xx

So,

1

352 2

x

xx1

472

xx

Division and the Remainder Theorem

)1( x )1( x

Solution:

e.g.2 Divide by352 2 xx 1x

1

352 2

x

xx1x

x2

• Correct the last term . . .

7 4

• Check the numerator.

So,

1

352 2

x

xx1

472

xx

• Write the denominator on the top line

• Correct the 1st term.

• Copy the denominator and correct the next term.

Division and the Remainder Theorem

)1( x

Solution:

e.g.3 Divide by32 23 xx 1x

1x

2x

33 xx

1

32 23

x

xx

Tip: As there is no linear x-term leave a space

Division and the Remainder Theorem

)1( x )1( xx

Solution:

e.g.3 Divide by32 23 xx 1x

1x

2x

1

32 23

x

xx

2222 xxx

Division and the Remainder Theorem

)1( x )1( x )1( x

Solution:

e.g.3 Divide by32 23 xx 1x

1x

2x

1

32 23

x

xx x 1

xx 0

xx )1(Be careful!

Division and the Remainder Theorem

)1( x )1( x )1( x

Solution:

e.g.3 Divide by32 23 xx 1x

1x

2x

1

32 23

x

xx x

213

21

Division and the Remainder Theorem

)1( x )1( x )1( x

Solution:

e.g.3 Divide by32 23 xx 1x

1x

2x

1

32 23

x

xx x 1 2

1

21

1

32 223

x

xxx

xx

The quotient is and the remainder is

212 xx

Division and the Remainder Theorem

)2( x )2( x 8

Divide by x + 2442 xx1.

Solution:

The quotient is x + 2 and the remainder is

8

The solution is on the next slide

Exercises

2

442

x

xx

2x

2

82

x

x

2

442

x

xx

2. Divide by143 23 xxx 1x

x 2

Division and the Remainder Theorem

)1( x )1( x )1( x

Solution:

1

143 23

x

xxx

1x

2x x2 6 7

1

762

1

143 223

x

xxx

xxx

The quotient is and the remainder is

7622 xx

Division and the Remainder Theorem

e.g. Find the remainder when is

divided by x - 1

47 23 xx

The remainder theorem gives the remainder when a polynomial is divided by a linear factorIt doesn’t enable us to find the quotient

47)( 23 xxxfLet 4)1(7)1()1( 23 f

4

The method is the same as that for the factor theorem

The remainder is 4

The remainder theorem says that if we divide a polynomial by x – a, the remainder is given by

)(xf )(af

Division and the Remainder Theorem

ax

Rxg

ax

xf

)(

)(

Proof of the Remainder theoremLet be a polynomial that is divided by x - a

)(xf

The quotient is another polynomial and the remainder is a constant.

We can write

Multiplying by x – a gives

Rxgaxxf )()()(

So, Ragaaaf )()()(

Raf )(

Division and the Remainder Theorem

e.g.1 Find the remainder when is divided by

143 23 xxx2x

Solution: Let

143)( 23 xxxxf

1)2(4)2(3)2()2( 23 f

)2(2 fRa So,

18128 13 R

Tip: Use the remainder theorem to check the remainder when using long division. If the remainder is correct the quotient will be too!

Division and the Remainder Theorem

e.g.2 Find the remainder when is divided by

4223 xxx12 x

Solution: Let

42)( 23 xxxxf

To find the value of a, we let 2x + 1 = 0. The value of x gives the value of a.

21012 xx

42212

213

21

21 f

4141

81

21 f

so, 21 fR

8

32821 R

8

21R

Division and the Remainder Theorem

Exercises Find the remainder when is

divided by x + 1 534 23 xxx1.

Solution: Let

534)( 23 xxxxf

5R5341 R

5)1(3)1(4)1()1( 23 f

2. If x + 2 is a factor of and if the remainder on division by x – 1 is – 3, find the values of a and b.

623 bxaxx

Division and the Remainder Theorem

Exercises

2. If x + 2 is a factor of and if the remainder on division by x – 1 is – 3, find the values of a and b.

623 bxaxx

06)2()2()2( 23 ba 0)2(f

3)1(f 361 ba

01424 ba

Solution: Let

6)( 23 bxaxxxf

2 ba - - - (2)

72 ba - - - (1)

(1) + (2)

393 aa

Substitute in (2)

1b

Division and the Remainder Theorem

Division and the Remainder Theorem

The following slides contain repeats of information on earlier slides, shown without colour, so that they can be printed and photocopied.For most purposes the slides can be printed as “Handouts” with up to 6 slides per sheet.

Division and the Remainder Theorem

Dividing by an expression of the form x - a can be done in 2 ways:

Method 1: Long Division

Method 2: Inspection

Method 2 is usually easier but an example of method 1 is given next.

Division and the Remainder Theorem

621 xx

Write the division as follows: 2

e.g.1 Divide by62 x 1x

• Multiply x – 1 by this number . . .

and write the answer below

• Subtract:6 – (– 2) = 8

8

The quotient is 2 and the remainder is 8.

So,1

82

1

62

xx

x

22 x)1(2 x

22 x

Solution: Method 1 Long division.

• Divide the 1st term of the numerator by the 1st term of the denominator.

• Write this answer above the polynomial being divided.

x2x 2

Division and the Remainder Theorem

1

x1

62

x

x 82 )1( x6

e.g.1 Divide by62 x 1x

• Write

• Adjust the constant term

1

8

1

)1(2

1

62

xx

x

x

x• Separate the 2 terms:

The quotient is 2 and the remainder is 8

Solution: Method 2 ( Inspection )

• Copy the denominator onto the top line• Divide the 1st term of the numerator

• Multiply

by the 1st term of the denominator

so the 1st term at the top is now correct

Division and the Remainder TheoremThe remainder theorem says that if we divide

a polynomial by x – a, the remainder is given by

)(xf )(af

ax

Rxg

ax

xf

)(

)(

Proof of the Remainder theoremLet be a polynomial that is divided by x - a

)(xf

The quotient is another polynomial and the remainder is a constant.

We can write

Multiplying by x – a gives

Rxgaxxf )()()(

So, Ragaaaf )()()(

R

Division and the Remainder Theorem

Solution: Let

143)( 23 xxxxf

1)2(4)2(3)2()2( 23 f

)2(2 fRa So,

18128 13 R

Tip: Use the remainder theorem to check the remainder when using long division. If the remainder is correct the quotient will be too!

e.g.1 Find the remainder when is divided by

143 23 xxx2x

Division and the Remainder Theorem

e.g.2 Find the remainder when is divided by

4223 xxx12 x

Solution: Let

42)( 23 xxxxf

To find the value of a, we let 2x + 1 = 0. The value of x gives the value of a.

21012 xx

42212

213

21

21 f

4141

81

21 f

so, 21 fR

8

32821 R

8

21R