[PDF] AMSG11Remainder and Factor Theorempdf

101353_6AMSG_11_RemainderandFactorTheorem.pdf www.faspassmaths.com11: THE REMAINDER AND FACTOR THEOREM

Solving and simplifying polynomials

nn our study of quadratics3 one of the methods used to simplify and solve was factorisation+ For example3 we may solve for x in the following equation as follows��� aence3 x ��� !. or !× are solutions or roots of the quadratic equation+ A more general name for a quadratic is a polynomial of degree ×3 since the highest power of the unknown is two+ The method of factorisation worked for quadratics whose solutions are integers or rational numbers+

For a polynomial of order .3 such as

the method of factorisation may also be applied+ aowever3 obtaining the factors is not as simple as it was for quadratics+ We would likely have to write down three linear factors3 which may prove difficult+ nn this section3 we will learn to use the remainder and factor theorems to factorise and to solve polynomials that are of degree higher than ×+ Before doing so3 let us review the meaning of basic terms in division+

Terms in division

We are familiar with division in arithmetic+

The number that is to be divided is called the dividend+ The dividend is divided by the divisor+ The result is the quotient and the remainder is what is left over.

From the above example, we can deduce that:

489 = (15 " .×- ���

! ! ! !

Dividend Quotient Divisor Remainder

Thus, from arithmetic, we know that we can express a dividend as: Dividend = (Quotient Divisor) + Remainder

When there is no remainder3

"#$, and the divisor is now a factor of the number, so

Dividend = Quotient Factor

The process we followed in arithmetic when dividing is very similar to what is to be done in algebra+ Examine the following division problems in algebra and note the similarities+

Division of a polynomial by a linear expression

We can apply the same principles in arithmetic to dividing algebraic expressions+ ���et the quadratic function %&'( represent the dividend3 and &')*(+the divisor3 where %&'(#,' - )'./

From the above example, we can deduce that:

,' - )'./#&,'./(&')*(.0 ! ! ! !

Dividend Quotient Divisor Remainder

Consider

& ')*(#$1 %&'(#&,'./(2$++.++0#0

But3 &

')*(#$ implies that '#*

Therefore, when

'#*, %&'(#0+ or %&*(#03

We can conclude that when the polynomial

,' - )'./+ is divided by & ')*(1 the remainder is %&*(#0

We shall now perform division using a cubic

polynomial as our dividend. x 2 +5x+6=0 (x+3)(x+2)=0 x+3=0,!x="3 x+2=0!x="2 x 3 +4x 2 +x!6=0

32 48932

15 169 160
9 ! ! x!1 3x 2 !x+23x+2 !(3x 2 !3x) 2x+2 !(2x!2) 4

Quotient

Dividend

Remainder

Divisor

Remainder

Dividend

Quotient

Divisor Copyright © 2019. Some Rights Reserved. www.faspassmaths.comPg 1 of 7 www.faspassmaths.com

From the above example, we can deduce that:

��� 4 +12��� - -3���+4=(' - .*0'./5(&')/(.50 ↑ ↑ ↑ ↑ Dividend Quotient Divisor Remainder

Consider

( ')/(#$1 ���('(#&' - .*0'./5(2$++.+50#50

But, (

')/(#$ implies that ���=2

Therefore, when

���=2, ���('(#50+ or ���(/(#503

We can conclude that when the polynomial

��� 4 +12��� - -3���+4 is divided by ( ')/(1 the remainder is ���(2)=54 Now consider another example of a cubic polynomial divided by a linear divisor.

From the above example, we can deduce that:

2���

4 -3��� - +4���+5=(/' - .'.*6(&'./(),* ↑ ↑ ↑ ↑ Dividend Quotient Divisor Remainder

Consider

( './(#$1 ���('(#&/' - .'.*6(2$+)+,*#),*

But, (

'./(#$ implies that ���=-2

Therefore, when

���=-2, ���('(#),*+ or ���()/(#),*3 We can conclude that when the polynomial

2���

4 -3��� - +4���+5 is divided by ('./(1 the remainder is ���()/(#),* The Remainder Theorem for divisor (���-���) From the above examples, we saw that a polynomial can be expressed as a product of the quotient and the divisor plus the remainder:

3���

- -���+2=(,'./(&')*(.0 ��� 4 +12��� - -3���+4=(' - .*0'./5(&')/(.50

2���

4 -3��� - +4���+5=(/' - .'.*6(&'./(),* We can now formulate the following expression where, is a polynomial whose quotient is and whose remainder is R when divided by .

If we were to substitute

���=��� in the above expression, then our result will be equal to ���, the remainder when the divisor, (���-���) is divided by the polynomial, ���('(3+

We are now able to state the remainder theorem

.

The Remainder Theorem

If is any polynomial and is divided

by , then the remainder is The validity of this theorem can be tested in any of the equations above, for example:

1.!When

3���

- -���+2 was divided by (')*(1 the remainder was 4. According to the remainder theorem, the remainder can be computed by substituting ���=1 in ���(���) ���('(#,' - )'./ ���(*(#,&*( - )*./ ���(1)=4

2.!When

��� 4 +12��� - -3���+4 was divided by ( ')/(1 the remainder was 54. According to the remainder theorem, the remainder can be computed by substituting ���=2 in ���(���) ���('(#' 4 .*/' - ),'.0 ���(/(#&/( 4 .*/&/( - ),&/(.0 ���(/(#6.06);.0 ���(2)=54 x!2x 3 +12x 2 !3x+4x 2 +14x+25 !(x 3 !2x 2 ) 14 x 2 !3x+4 !(14x!28x) 25x+4 !(25x!50)
54
x+2 2x 3 !3x 2 +4x+52x 2 +7x+18 !(2x 3 +4x 2 ) !7x 2 +4x+5 !(7x 2 !14x) 18x+5 !(18x+36) !31 f(x) Q(x) (x\a) () () ( )fx Qx x a R=!"+ f(x) f(x) (x\a) ().faCopyright © 2019. Some Rights Reserved. www.faspassmaths.comPg 2 of 7 www.faspassmaths.com

3.↑When

/' 4 ),' - .0'.5 was divided by &'./( the remainder was ),*. According to the remainder theorem, the remainder can be computed by substituting '#)/ in %&'( %&'(#/' 4 ),' - .0'.5 %&)/(#/&)/( 4 ),&)/( - .0&)/(.5 %&)/(#)*;)*/)6.5 %&)/(#),*

Example 1

Find the remainder when

%&'(#,' 4 .' - )0')* is divided by &')/(.

Solution

By the Remainder Theorem, the remainder is

%&/(3 %&'(#,' 4 .' - )0')* %&/(#,&/( 4 .&/( - )0&/()* %&/(#/0.0)6)* %&/(#*<

Hence, the remainder is 19

The Factor Theorem for divisor

&7)8( Now, consider the following examples when there is no remainder. We can express the dividend as a product of the divisor and the quotient only, since :#$. %&'(#,' - )')/#&,'./(&')*( %&'(#6' 4 )*$' - )'.,#&6' - )/'),(&')*( We can now formulate the following expression where %&'( is a polynomial, =&'( is the quotient and &')9( is a factor of the polynomial. %&'(#=&'(2&')9( The above rule is called the Factor Theorem, it is a special case of the Remainder Theorem, when :#$. The validity of this theorem can be tested by substituting '#* in each of the above functions. %&'(#,' - )')/ %&*(#,&*( - )*)/ %&*(#$ %&'(#6' 4 )*$' - )'., %&*(#6&*( 4 )*$&*( - )* ., %&*(#$

In the above examples, when we let &

')*(#$, or '#*, %&'(#$++because the remainder, :#$.

The Factor Theorem

If is any polynomial and is divided

by , and the remainder then is a factor of

Example 2

Show that &')/( is a factor of

%&'(#,' 4 .' - )*0'

Solution

By the Factor Theorem, if

&')/( is a factor of the remainder is zero. We now compute the remainder, %&/(3 %&'(#,' 4 .' - )*0' %&/(#,&/( 4 .&/( - )*0&/( %&/(#/0.0)/6 %&/(#$

Hence,

&')/( is a factor of %&'(3

The Remainder and Factor Theorem for

divisor &87.>( When the divisor is not in the form, , but in the general linear form &9'.?(, the remainder can no longer be %&9(3 This is because the coefficient of x is not equal to one. Consider the following example, where the divisor is of the form, &9'.?(. Let &/'.,( be a divisor of %&'(#/' 4 .@' - ./'.<

We perform the division as shown below and note

that the remainder is 15. x-1 3x 2 -x-23x+2 -(3x 2 -3x) 2 x-2 -(2x-2) 0 x\1 8x 3 \10x 2 \x+38x 2 \2x\3 \(8x 3 \8x 2 ) \2x 2 \x+3 \(\2x 2 +2x) \3x+3 \(3x+3) 0 f(x) f(x) (x!a) () 0fa= (x!a) f(x) (x!a)Copyright © 2019. Some Rights Reserved. www.faspassmaths.comPg 3 of 7 www.faspassmaths.com

We can deduce that:

2���

4 +7��� - +2���+9=(' - ./')/(&/'.,(.*5

Consider (/'.,(#$1

���('(#&' - ./')/(2$++.+*5#*5

But, (

/'.,(#$ implies that ���=- 4 - .

Therefore, when

���=- 4 - , ���('(#*5+ or ���A) 4 -

B#*53

We can conclude that when the polynomial

2���

4 +7��� - +2���+9 is divided by (/'.,(1 the remainder is ���A) 4 - B#*5

Now consider the example below.

We can deduce that:

9���

4 +15��� - -9���+1=(,' - .;')*(&,')*(

Consider (,')*(#$1

���('(#&,' - .;')*(2$++#$

But, (

,')*(#$ implies that ���= C 4 .

Therefore, when

���= C 4 , ���('(#$+or ���A C 4 B#$3

We can conclude that when the polynomial

9���

4 +15��� - -9���+1 is divided by (,')*(1 the remainder is ���A C 4 B#$3 So, (3���-1) is a factor of 9��� 4 +15��� - -9���+1 From the above examples, we can formulate the following expression: ���('(#=&'(2&9'.?(.: where, is a polynomial whose quotient is and the remainder is R when divided by ( ������+���).

By setting

(

9'.?(#$1 the above polynomial will

become ���('(#: When (

9'.?(#$1 ���=-

D E .

Now, since

���('(#:, when ���=- D E , we conclude that ���F) ? 9 G#: We are now in a position to restate the remainder theorem when the divisor is of the form .

The Remainder Theorem

If is any polynomial and is divided by

then the remainder is .

If = 0, then is a factor of .

We apply the Remainder Theorem to obtain the

remainder when ���('(#/' 4 .@' - ./'.< was divided by (/'.,(3+

By the Remainder Theorem, the remainder is

���A) 4 - B3 ���F) , /

G#+/&)

, / ( 4 .@&) , / ( - ./&) , / (.< ���F) , / G#) /@ 0 . ;, 0 ),.< ���F) , / G#*5

We can apply the Factor Theorem to show that

( ,')*( is a factor of ���('(1 where ���('(#<' 4 .*5' - )<'.*+ By the Factor Theorem (,')*( is a factor of ���(���). if ���A C 4

B#$1

���('(#<' 4 .*5' - )<'.* ���F * , G#Hence, (,')*( is a factor of ���(���). Instead of performing long division, we can apply the remainder theorem to find the remainder when a polynomial is divided by a linear expression of the form ( ax + b). The remainder, ��� =���A) D E

B+when the

polynomial is divided by the linear factor.

2x+3 2x

3 +7x 2 +2x+9x 2 +2x!2 !(2x 3 +3x 2 ) 4 x 2 +2x+9 !(4x 2 +6x) !4x+9 !(4x!6) 15

3x\1 9x

3 +15x 2 \9x+13x 2 +6x\1 \(9x 3 \3x 2 ) 18x 2 \9x+1 \(18x 2 \6x) \3x+1 \(\3x+1) 0 f(x) Q(x) ( )ax b+ ( )fx( )fx ( )ax b+ bfa!"#$%&' bfa!"#$%&' ( )ax b+( )fxCopyright © 2019. Some Rights Reserved. www.faspassmaths.comPg 4 of 7 www.faspassmaths.com We can use the factor theorem to show that a linear expression of the form ( ax + b) is a factor of a polynomial. In this case, we show that the remainder :+#%A) D E

B+is zero when the polynomial is divided

by the linear factor.

Example 3

Find the remainder when

is divided by .

Solution

In this example is of the form ,

where and b = -1

The remainder would be

. Alternatively, we could have used long division to show that the remainder is one half.

Example 4

State the quotient and the remainder when

;' 4 )6'.5 is divided by by /')0+

Solution

In this example, the dividend has no terms in

' -

3 It is

advisable to add on such a term to maintain consistency between the quotient and the divisor. This is done by inserting a term in ' - as shown below.

The quotient is

,' - .;'.6

The Remainder is 37

Factorising and Solving Polynomials

We can use the factor theorem to factorise or solve a polynomial. However, to factorise a polynomial of the form it would be helpful to know one linear factor. Then we can obtain the other factors by the process of long division.

Example 5

Show that (2x + 3) is a factor of

.

Solution

When f(x) is divided by , the remainder is

Hence, is a factor of since the

remainder is 0.

Example 6

Factorise, %&'(# and hence

solve .

Solution

Let .

To obtain the first factor, we use the remainder

theorem to test for f(1), f(-1) and so on, until we obtain a remainder of zero.

We found that,

%&0(#$

Therefore (

x Ð 4 ) is a factor of f(x).

Now that we have found a first factor, we divide

f(x) by ( x - 4). ( ) 32

4231fx x x x=+!+( )21x!

( )21x!( )ax b+ 2a= ( )11

22ff!!"#"#=$%$%&'&'

32

11 1 1423122 2 2

11 1 11122 2 2f

!" !" !" !"=+#+$% $% $% $%&' &' &' &' =+#+= 21
2 32
32
2 2 1 2 1 2 2 2

214 2 31

42
43
42
1 xx xxxx xx xx xx x x+! ! +!! !! ! !! ! + !!+

2x!4 6x

3 +0x 2 !8x+53x 2 +6x+8 !(6x 3 !12x 2 ) 12 x 2 !8x+5 !(12x 2 !24x) 16 x+5 !(16x!32) 37
32
ax bx cx d+++ ( ) 32

2323fx x x x=+!!

( )23x+

3233 3 3

22 2 2

()2()3()2()30f!=!+! !!!= ( )23x+( )fx 32

2 3 29 60xx x+!!

( )0fx= 32
() 2 3 29 60fx x x x=+-- 32

(4) 2(4) 3(4) 29(4) 60 128 48 116 60 0f=+--=+--=Copyright © 2019. Some Rights Reserved. www.faspassmaths.comPg 5 of 7

www.faspassmaths.com

If , then .

So ���=-3,2

C - and 4.

Example 7

Factorise and hence show that

has only one solution.

Solution

To get the first factor, we apply the remainder

theorem as follows:

Hence, is a factor of .

Dividing by the factor

The quadratic has no real roots because

the discriminant

Hence, has only one

solution,

Example 8

If and are both factors of

, find the third factor.

Solution

Let the third factor be

(������+���). We can write the expression as a product of three linear factors as shown. And

Hence, ( ), the third factor, is .

Alternatively, we may divide

by .

The third factor is .

Example 9

Solve for x in , giving the

answer to 2 decimal places where necessary.

Solution

Recall: If is any polynomial and is

divided by , then the remainder is .

If the remainder , then is a factor

of .

First test to see if (���-1) is a factor.

Let is not a factor of .

Next, test to see if (���+1) is a factor.

2 32
32
2 2 2 11 15

42 3 29 60

2 8 11 29 11 44 15 60 15 60 0xx xxxx xx xx xx x x++ ! +!! !! ! !! ! !! ( )( ) 2

21115 325xx x x++=+ +

( ) ( )( )( )25 3 4fx x x x!=++" ( )0fx=( )( )( )25 3 40xxx++!= 32

4 12xx x!+!

32

4 12 0xx x!+!=

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 32
32
32

1 1 1 4 1 12 0

1 1 1 4 1 12 0

2 2 2 4 2 12 0f

f f=!+!" ! =! !!+!! " = !+!= (2)x\()fx (2)x! 2 32
32
2 2 6

2 4 12

2 4 12 2 6 12 6 12 0xx xxxx xx xx xx x x++ !!+! !! + ! !! ! !! 32
2 2 412
(2)( 6) (2)0 or ( 6)0

2 xx x

xxx xxx x-+- = -++ "-= ++= " = 2 6xx++ 22

4(1)4(1)(6)125240bac!=!=!=!<

32

4 12 0xx x!+!=

2x= ( )3x+( )4x- 32

232960xx x+--

( )( )( ) 32
3

232960 3 4

2

2xx x x x axb

xxax x a+--=+-+

´´=

\ = 34 60
5b b´- ´=- \ = ax b+ ( )25x+ ( )( ) 2

34 12xx xx+-=--

32

2 3 29 60xx x+--

2

12xx--

232
32
2 2 25

12 2 3 29 60

2224
5560
5560
0x xx x x x xx x xx xx+ --+- - --- -- --- \ ( )25x+ 32

2 7 12 0xxx--+=

( )fx( )fx ( )xa-( )fa ( )0fa=( )xa- ( )fx ( ) 32

2712fx x x x=- -+

( ) ( ) ( ) ( ) 32

1 1 2 1 7 1 12 0f=- -+¹

( )1x\-( )fxCopyright © 2019. Some Rights Reserved. www.faspassmaths.comPg 6 of 7 www.faspassmaths.com is not a factor of .

Now, try (���-2)

is not a factor of .

Try (���+2)

is not a factor of .

Try (���-3)

f ( 3) = (3) 3 Ð 2(3) 2 -7(3) +12 = 0 is a factor of .

Now we divide by .

Since we were asked to give our answers correct to

2 decimal places, we deduce that = 0

would not have exact roots. We, therefore, employ the quadratic equation formula to find the roots.

If , then ,

where ���=1,���=1 and ���=-4.

Hence, .

( ) ( ) ( ) ( ) 32

1 1 2 1 7 1 12 0f-=- -- --+¹

( )1x\+( )fx ( ) ( ) ( ) ( ) 32

2 2 2 2 7 2 12 0f=- -+¹

( )2x\-( )fx ( ) ( ) ( ) ( ) 32

2 2 2 2 7 2 12 0f-=- -- --+¹

( )2x\+( )fx ( )3x\-( )fx ( )fx( )3x- 2 32
32
2 2 4 32712
3 712
3 412
412
0xx xxxx xx xx xx x x+- ---+ -- - + -- - + --+ 2 4xx+- 2

0ax bx c++=

2 4

2bb acxa-±-=

( ) ( ) ( )( ) ( ) 2 11414
21
117
2

1.561 or 2.561

1.56 or 2.56 (correct to 2 decimal places)x

-

±- -=

- ± = = - = -

3, 1.56 or 2.56x=-Copyright © 2019. Some Rights Reserved. www.faspassmaths.comPg 7 of 7

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