The graph suggests that the function has three zeros, one of which is x = 2 It's easy to show that f(2) = 0, but the other two zeros seem to be less
In this section, we will learn to use the remainder and factor theorems to factorise and to solve polynomials that are of degree higher than 2 Before doing so,
Factor Theorem : If p(x) is a polynomial of degree n > 1 and a is any real number, then (i) x – a is a factor of p(x), if p(a) = 0, and (ii) p(a) = 0, if x –
Class 9 Maths Polynomials Factor Theorem Factor Theorem x a hook a factor of the polynomial px if pa 0 Also if x a passage a factor of px then pa 0 where
Study more about this topic with online classes for 9th Class Maths Remainder Theorem: Let p(x) be any polynomial of degree greater than or equal to one and
2 sept 2021 · Through examples interpret to find Discuss and recall about definition, CLASS:9 POLYNOMIALS ACTIVITY SHEET-07 FACTOR THEOREM
3 5 3 5 Factor P x x x x x x x x x x = + + ? = + + ? Page 18 Example 9 (Continued): From the Rational Zeros Theorem (Section 5 3), we obtain the
Class: 9 By: Manish Gupta polynomial p(x) thus we define the zero of a polynomial as follows 05 Division of Polynomials and Remainder Theorem
22 avr 2020 · https://ncerthelp com/text php?ques=Polynomials+Class+9+Maths+ degree of zero polynomial is not defined Remainder theorem : Let
According to the remainder theorem, p(x) divided by (x-1) obtains the remainder as g(1) Calculating g(1) = 1 3 ? 6(1)2 + 9 × 1 + 3
101390_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