13.

1 REVIEW
FACTORISATION
13
Factors
Each of the numbers (constant or variable), that forms a product is called
a factor of the product.
(i) 5 and x are factors of product 5x.
(i) (2x- 5) and (3x +2) are the factors of (2x-5) (3x +2).
Since (2x- 5)(3x + 2) =2x(3x +2) -5(3x +2)
=6x+ 4x- 15x -10
=6X2-11x-10
.2x-5 and 3x +2are factors of 6x-11x- 10,

13.2 | F A C T O R I S A T I O N

product is equal to the

Factorisation means to find two or more expressions whose
given expression.

13.3FACTORISATION BY TAKING OUT COMMON FACTORS

monomial that will divide each term of the
Steps :1. Find by inspection, the largest
given polynomial completely.
this monomial (factor) and
2. Divide each term of the given polynomial by common monomial outside
enclose the quotient within brackets keeping
this
the bracket.

Example 1:
Factorise: () 5x2- 10x (ii) 3x²y - 6xy² + 9xy
Solution:
largest monomial which divides each
term of the given
() By inspection,we find that the (Step 1]
polynomial 5x- 10x is 5x.
5x 10x (Step 2)
5x2-1Ox =5x 5x
5x
(Ans.)
= 5x (x - 2)
3x²y 6xy 9xy
(0) 3x²y - 6xy² + 9xy = 3xy 3xy 3xy 3xy
(Ans.)
= 3xy (x-2y + 3)
Example 2:
Factorise :
() -10ay? 15a6y + 20a'x5 (i) 2x(a + b) -3y (a + b)
Solution : 15ax4 +20a'x

10ax2
5atx* - 5ax2 - 5ax2
0 -10ax?- 15a6y + 20a x5 = - -5ax2

3a2y?- 4ax)
(Ans.)
= -5ax(2 +
169
Factorisation
 (ii)
2x(a + b) 3y(a +b)
2x(a + b) - 3y(a + b) = (a + b) a +b a +b
= (a+ b)(2x - 3y)
(Ans)
13.4FACTORISATION BY GROUPING
Agiven algebraic expression, containing an even number of terms
into factors, if its terms can be arranged in groups such that each group has amay be resolved
Steps : 1. Arrange the terms of the given expression in suitable common factor.
each group has a common factor.
groups sSuch that
2. Factorise each group.
3. Take out the factor which is common to each group.
Example 3:
Factorise: ax -bx + ay - by
Solution:
ax - bx + ay- by
= (ax - bx) + (ay - by)
= x(a- b) + yla - b) (Step 1]
= (a b)(x + y) IStep 2)
OR ax- bx + ay - by
(Ans.) [Step 3)
=ax + ay - bx - by
= a(x + y) -b(x + y) [Step 1]
= (X + y)a b) [Step 2
Example 4: (Ans.) [(Step 3
Factorise :
() y-3y² + 2y - 6- xy + 3x (ii) a?- (b + 5) a + 5b
Solution:
(i) y-3y² + 2y -6- xy + 3x = (y-3y') + (2y - 6) -
(xy 3x) [Step 1]
y²(y - 3) + 2(y - 3) - x(y - 3) (Step 2]
= (y-3)(y² +2 -x)
(Ans.) (Step 3)
(i) a- (b + 5) a + 5b = a²- ab -5a+ 5b
[Removing the bracket
= (a- ab) (5a- 5b)
(Step 1]
= a(a- b) -5(a - b)
[Step 2]
= (a- b) (a - 5)
(Ans.) (Step 3)

1. Multiple Choice Type :

EXERCISE 13(A)
Choose the correct answer from the (a) (x- y)² (a - b)
given below. options (b) (X + y) (x- y) (a- b)
(() -7x- 14y is equal to : (c) (x- y) (a -bx + by)
(a) -21x²y (d) (x-y) (a - bx - by)
(b) 14x²y (ii) a + bc + ab + ac is equal to:
(c) 7(-x?-2y) (d) -7(x2+2y)
(ii) a(x - y) - b(x - y) is equal to : (a) (a + b) (a + c) (b) (a+ b) (b +c)
(c) (a + c) (a - b) (d) (a-b) (0 +
170
Concise MATHEMATICS Middle Schoo
 10. 6xy(a² +b²) +8yz(a² +b)-10xz(a? +b)
1-2x-2x2. +4x°is equal to :
) (1-2x2) (b) (1 +2x)
(1 +2x2) 11. xy - ay - ax + a + bx - ab
+2x) )
(a) (1 (1--2x2) (d) (1-2x)
(1 +2x2) 12. 3x5 6x-2x3 + 4x? +X-2
(1-2x)
(c) :
-y) - b(y - x) is equal to 13. x²y -X+3xy +3
Ma(x
- by + bx) 14. 6a?-3ab - b² + 202
(a) (x-y) (a bx + by)
- 15. 3a²b - 12a²- 9b + 36
(b) (y-x) (a
(a- bx - by) 16. x2- (a -3)x 3a
(c) (x-y)
+ by)
(d) (x-y)(a- bx 17. ab²- (a-c) b-C
51a2%4
17ab- 34a b® + 18. (a2- b) c +(b²-c)a
2
+ 12x3s
3rfy-27xty²
19. a- a2- ab + a+b-1
3
- b) +z'la - b)
4 sla-b) -yla y)la+ b) 20. ab(c² +d²)- a?cd - bcd
(*-
5 (K+yla +b) + 21. 2ab² - aby + 2cby - cy²
3c (2a + b)
6. 2b (2a +b) - 3ab33
6a?bc2 + 22. ax + 2bx + 3cX - 3a- 6b-9c
7. 12abc -
8. 4x(3x-2y) -
2y(3x -2y) 23. 2abc-2a + 3bc -3b 4bc2+ 4c
(a+2b) (3a + b) - (a + b) (a + 2b) + (a + 2b)2
9.

H35FACTORISATION OF DIFFERENCE OF TWO SQUARES

(x- y) is x²- y²
Since the product of (x + y) and
Factors of x²- y² = (x+ y)(X - y)
their difference.
squares of two terms = Sum of the two terms x
Difference of

Example 5 :
Factorise: 25a2 - 36b2
Solution : 6b)(5a - 6b) (Ans.)
25a? - 36b² =(5a)2- (6b) = (5a +
Example 6:
Factorise : (0) 1-4(a -2b) (i) 9(x + y)- 16(x-3y)2
Solution :
1-4(a -2b)² = 1-2 (a -2b)
= 1- [2(a - 2b)]
= 12- (2a 4b)
- 2a 4b)
= (1+ 2a-4b ) (1 (Ans.)
4b)
= (1 +2a - 4b)(1-2a+
3y))
) 9(x +y)?-16(x -3y)² =[3(x + y)-[4(x-
12y)2
= (3x+ 3y)² (4x -
-12y )
(3x +3y + 4x- 12y) (3x +3y - 4x
= 12y)
12y)(3x + 3y - 4x +
= (3x + 3y + 4x - (Ans.)
= (7x 9y)(15y -x)
171
 EXERCISE 13(B)
(b) (a +1+b-x) (a
1. Multiple Choice Type : (c)
+1-b+x)
Choose the correct answer from
the options (a-1+b- x) (a -1-b+x
given below. (d) (a-1+ bx) (a+1- bx)
(1) (2x + y)² - (2y + x)² is equal to : 2. (a + 2b)2 a2
(a) 3(x + y) (x- y) (b) 2(x - y) (x + y) 3. (5a -3b)2- 16b2
(c) 2(y - x) (x + y) (d) (3x + y) (3x - y)
4. a- (a-3b2)2
(ii) 49- (x + 5) is equal to: 5. (5a - 2b)2- (2a -b)2
(a) (54 - x) (54 + x) (b) (2 -x) (12 + x)
(c) 48(x + 5)2 (d) 48(x - 5)2 6. 1-25 (a +b)²
(iüi) a-2ab + b² +a-b is equal to : 7. 4(2a+ b)² (a- b)?
(a) (a - b) (a + b- 1) (b) (a-b) (a +b+1) 8. 25(2x + y)- 16(x - y)2
(c) (a + b)(a b- 1)(d) (a-b) (a-b+ 1)
(iv) x² + y² - 2xy -1 is equal to :
(a) (x + y- 1)(x - y- 1)
9.
(*3)-("4)
10. (0-7)2- (0-3)2
(b) (x + y + 1)(x - y- 1) 11. 75(x + y)² 48(x - y)2
(c) (x + y+ 1) (x -y+ 1) 12. a² + 4a + 4- b2
(d) (x- y+ 1) (x - y-1) 13. a²- b²2b 1
(v) a² + 2a + 1- b²- x² +2bx is equal to : 14. x² + 6x +9- 4y²
(a) (a + 1-b+ x) (a -1-b+x)

13.6FACTORISATION OF TRINOMIALS
Since the product of two binomials (2a + b)and (3a - 5b)
= (2a + b)(3a - 5b)
= 6a?-7ab - 5b2, which is a trinomial.
The factors of a trinomial 6a²-7ab - 5b² are the binomials (2a + b) and (3a-5),
Before learning the factorisation of a trinomial, it is essential to know how to find out
the two numbers whose product and sum are given.
Example7:
Find the numbers whose :
() product =6 and sum =5 (ii) product = 6 and sum =-5
(ii) product = 6 and sum = 5 (iv) product =-6 and sum =-5
Solution:
() Since product = 6 and sum =5. The product and the sum of two numbers are pos0e
only when both the numbers are positive.
By trial, we find that the required two numbers are 3 and 2. (Ans)

Product of 3 and2=3x 2=6 and their sum = 3+ 2=5

(ii) Since product =6 and sum =-5
The product of two numbers is positive and their sum is negative only when boththe
numbers are negative.
(Ans)
Required numbers are -3 and- 2.

Schoo-8
 product = 6 and sum = 5.
) Since,
product
is negative and their sum is positive only when the larger
of two numbers
The
two n
numbers is positive and the smaller is negative.
ofthe
we find that the
required two numbers are 6and-1. (Ans.)
By trial,
product -6 and sum =-5. The product of two numbers is negative and their
lw)
Since
also negative only
when the larger numberis negative andthesmalleris positive.
Sumis

we find that therequired

two numbers are - 6 and 1. (Ans.)
. By trial,
ofa trinomial
are :
standard forms
The
Note:
6x+11x
+3 i.e. descending order of the powers of its literal coeficients.
(0 coefficients.
6x2 ie. ascending order of the powers of its literal
3+11x +
given trinomial, the following steps
should be adopted :
factorise a
To the first and the last terms of the trinomial with their signs. In case
product
1. Findthe terms = 6x² x 3= 18x2,
ftinomial 6x² + 11xX +3, the product of its first and last
the sum of these two
A Golit the middle term of the given trinomial (.e. + 11x) such that
product obtained in
tarrns is equal to the middle term and their product is equal to the
step 1 (.e. 18x2)
and+2x.
By trial, we find thatthe twO Such terms are + 9x
given trinomial,.
9. Now by forming the suitable groups, factorise the
ie. 6x2+ 11x +3 = 6x + 9x + 2x +3
= 3x(2x + 3) + 1(2x +3)
= (2x + 3)(3x + 1) (Ans.)
Example 8:
Factorise:
(0 x'-9x+ 20 (i) y² +5y- 24 (ii) 1-3a - 28a2
Solution :
(0) Given trinomial = x²9x + 20
The product of its first and the last terms = x²x 20 = 20x?
Splitting the middle term (i.e. -9x) into two terms so that their product is 20x and
sum
is -9x, we get: -5x and-4x.
X-9% + 20 = X*-5x 4x + 20
= x(x- 5) -4(x-5)
= (X-5)(x 4) (Ans.)
(0) Given trinomial is y² + 5y-24
The product of its first and the last terms =y²x-24 = - 24y'
and the middle term = +5y.
NOW Tind two terms whose product should be -24y² and sum should be + 5y. By trial,
We find that the reguired two terms are + 8y and- 3y.
y'+ 5y-24 = y² + 8y - 3y -24
=yly + 8)-3(y + 8) = (y + 8)(y -3) (Ans.)
Facdoisation 173
 (ii) Given trinomial is 1- 3a -28a
Product of the first and the last terms = 1x-28a = -28a2
and the middle term =-3a
Bytrial, wefind thattwo terns whose product is -28a? and sum is-3a:
1-3a - 28a? = 1-7a + 4a- 28a?
= 1(1-7a) + 4a(1 -7a) =(1 -7a)(1 + 4a)
are-7aand +44
Example9:
Factorise : () (a +b)- 11(a +b) - 42 (ü) 7+ 10(x -y)
Solution: 8(x- y?
() (a + b)?- 11(a + b) - 42 = x²-11x- 42
=X-14x + 3x- 42 Taking a+b
= x(x- 14) +3(x 14) (Splitting the midde le
= (x-14)(x + 3)
= (a +b-14)(a + b+3) (Ans.)
(i) 7+10(x - y)- 8(x - y = 7+ 10a - 8a? ISubstuting x=a +
= 7+ 14a 4a -8a2 [Taking x-ya
= 7(1 + 2a) - 4a(1 + 2a) [Splitting the middle lem
= (1+ 2a)(7 4a)
= [1 + 2(x - y)l7 4(x- y)]
= (1 + 2x - 2y)(7 - 4x + 4y) [Substituting a=x-}
(Ans)
|13.7FACTORISING APERFECT SQUARE TRINOMIAL
Square of a binomlal is called a perftect square
Since, (a + b)² = a + 2ab + b² trinomial.
and, (a - b)² = a²-2ab + b2
a+ 2ab + b² and a? -2ab + b²
are perfect square trinomials.
Any trinomial which can be
square trinomial. expressed as a + 2ab + b or a-2ab +
b² is a pefect
Example 10:
(0) ls 4x + 12xy + 9y² a
perfect square trinomial ?
(i) Is x² - 6xy + 36y² a perfect
square trinomial?
Solution:
() 4x? + 12xy+9y² = (2x) +2 x 2x x3y + (3y)2
= a' +2ab + b²
=(a + b)2 Taking 2x =aand 3y =
= (2x + 3y)2
i. The given trinomial 4x² + (Ans)
12xy +
(ii) x²-6xy +36y² = ()² xx6y +9y2(6yis a perfect square trinomial.
= a-ab + b?
[Taking x= a and 6y =
Since the given
square trinomial. trinomial cannot be
expressed as a² - 2ab + b², it is not a pertet
(Ans!
174
Schoa-
Concise MATHEMATICS Middle
 EXERCISE 13(C)
Type :
Choice 2. a²+ 5a +6 3. a-5a + 6
. Multiple
correct
answer from the options
Choose
the 4. a+ 5a-6 5. x2+5xy +4y?
g i v e nb e l o w .

equal to : 6. a2-3a 40 7. x2-X-72

0x-9x -10 is
(b) (X- 10) (X-1) 8. 3a-5a +2 9. 2a-17ab + 26b2
la) (X-10) (X+1) (d) (x+ 10) (x+ 1) 11. 4c2 + 3c- 10
i) (x+ 10) (X-1) 10. 2x² + xy- 6y²
x2-23x + 42 is equal to : 12. 14x2+X-3 13. 6+ 7b-3b2
) (b) (x-21) (x -2)
a) (x-21) (x +2) 14. 5 +7x-6x2 15. 4+y- 14y²
(c) (x+21) (X+2) (d) (x+21) (x-2) 16. 5 +3a -14a2
(2x - 1) is equal to
:
6i) (482-4x + 1)+ (b) 2x -1 17. (2a+ b)² + 5(2a + b) + 6
(a) 2x+
1
(c) 2X- 1
(d) none of these 18. 1-(2x + 3y) 62x +3y)²
-4 is equal to : 19. (x-2y) - 12(x -2y) + 32
(v) (x +y)- 3(x + y)
y- 1) 20. 8 + 6(a + b) -5(a + b)²
(a) (x +y+4) (x+
1)
(b) (X +y +4) (* + y + 21. 2(x + 2y)2- 5(x +2y) +2
+ 1)
(c) (x+ y-4) (X + y 22. In each case, find whether the trinomial is a
(d) (x+y-4) (x + y - 1) perfect square or not:
to : (i) a²-10a + 25
(M) 60+ 11x - x* is equal (i) X+ 14x + 49
(15- x) (iv) 9b² + 12b + 16
(a) (4 +x) (15 - x) (b) (4-x) (iii) 4x2+ 4x + 1
() (4+ x) (15 - x) (d)
(4 + x) (15 + x*) (v) 16x2- 16xy + y² (vi) x²- 4x + 16

13.8 FACTORISING COMPLETELY

Example 11:
(ii) 3x² + 12x36
Factorise completely: (i) 8xß- 18xy2
Solution : (Taking out the common]
8x° - 18xy² = 2x (4x? 9y²) [Converting in the form a' - b)
=2x [(2x)²- (3y)9]
(Ans.)
= 2x (2xX +3y)(2x-3y)
[Taking out the common factor]
() 3x2 + 12x 36 = 3 (x² + 4X- 12) [Factorising the trinomial]
= 3 (x* + 6x-2X -12)
= 3 [x(x + 6) -2(x + 6)] (Ans.)
= 3(x+ 6)(x -2)
Example 12 :
Factorise completely : () x²+ 4xy + 4y² 97?
(i) 16x-y
Solution :
expression, x² + 4xy + 4y is a perfect square trinomial as:
9mne given
X*+ 4xy + 4y2 = x?+2xxx2y +(2y)* [Taking X= aand 2y = b]
= a²+2ab + b²
= (a + b)² [Substituting]
= (x + 2y)?
175
Faciorisation
 K+ 2y) (3z)²
= (X + 2y + 3z)(x + 2y - 3z)
(i) 16x* - y = (4x2)2 - (y)°
=(4x2 + y°)(4x² - y²) (Ans)
= (4x² + yI(2x)² - (y)
= (4x² + y°X2x + y(2x - y)

EXERCISE 13(D) (Ans.)

1. Multiple Choice Type :
Choose the correct answer from (b) 2x(3 -2x) (3 +
given below. the options 2x)
(c) 2x(2x + 3)
(i) x- 4x is equal to : (2x3)
(d) x(4x + 6y) (4x -
(a) x(x + 4) (x - 4) (b) (v) x2- (a b)x - 6y)
x(x + 2) (x -2) ab is equal to :
(c) (« + 4) (x 4) (d) (x+ 2) (x (a) (x- a) (x b)
(i) x-y + x²- y² is -2) (b) (X + a) (x-
equal to : (c) (x- a) (x + b) b)
(a) (x +y+ 1) (x + y - 1) (d) (x + a) (x +
(x*+ 2. 8x²y - 18y b)
(b) (x + y) (x- y) (x² + y²-1) y²) 3. 25x3-x
(c) (x + y) (x - y) (x?+ y² + 1)
4. 16x 81y 5. x²--3x -3y
6.
(d) none of these x2-y² -2x + 2y 7. 3x² + 15x-
(iil) x3-X* + ax + 8. 2a2 -8a 64 72
X-a-1 is equal to : 9. 3x²y + 11xy +
(a) (x- 1) (*²+ a -1) 10. 5ap² + 11ap + 6y
2a 11. a + 2ab
(b) (x- 1) (*² + a + 1) 12. x²+ 6xy + 9y² + +b²
X+ 3y
(c) (x- 1) (x²- a+ 1) 13, 4a?- 12ab +
9b² + 4a - 6b
(d) (x- 1) (x- a- 14. 2a?b2 - 98b
(iv) 8x3- 18x 0s equal 1) 15. a2- 16b2 -2a
to: - 8b
(a) x(2x + 3) (2x - 3)

1. Multiple Choice Type: Test yourself

Choose the correct answer from the
glven below. options (b) (x-2y) (x 2y + 3)
() (a + b)² 4ab is (a) (x +2y-3) (x + 2y)
equal to : (d) (x 2y) (*- 2y - 3)
(a) (a +b+ 2ab) (a +
b-2ab) (v) a(x - y) - by+ bx is
(b) (a + b) (a b) equal to
(c) (a + b) (a + b) (a) (x- y) (ax + by + b)
(d) (a - b) (a -b) (b) (X -y) (ax + by - b)
(i) a + 4a -32 is (c) (x- y) (x + y + a-b)
equal to :
(a) (a² + 8) (a + 2) (a + (d) (x - y) (ax - ay + b)
2)
(b) (a² -8) (a -2) (a + (vl) Statement 1: The
2) product
(c) (a + 8) (a² + 4) binomials is a trinomial, converselyof iftwo
we
(d) (a² + 8) (a + 2) (a factorise a trinomlalwe always obtaln two
(ii) 36 - 60y + 25y² ls -2) binomial factors.
equal to : Statement 2: The square of the difference
(a) (3 + 5y) (3 + 5y) (b) of two terms = The sum of the same WO
(c) (3 + 4y) (3 - 4y) (d) (3- 5y) (6 - 5y) terms.x thelr dlfference.
(iv) (X- 2y) - 3X + 6y is none of these
equal to : Which of the following optlons is correctr
(a) (x- 3y) (x + 2y) (a) Both the statements are true.
(b) Both the statements are false.
176
Concise MATHEMATICS Middle School
 Statement 1is true, and statement 2is (iv) ab(x² +y) -xy(a? +b)
false (v) m-1- (m-1)' + am
1 is false, and statement 2 -a
(d)
Slatement 4. Factorise :
is true.
questions are Assertion-Reason
(() 25(2x - y - 16(x-2y)²
tollowing (i) 16(5x +4) - 9(3x-2)2
The Choose your answer based
based i
questions.
below. (i) 25(x -2y) -4
given
a the codes and R are correct, and R is the 5. Factorise :
A
(1) Both for A. () a-23a + 42.
correct explanation (iü) a- 23a - 108
and Rare correct, and Ris not the (i) 1- 18x 63x? (iv) 5x- 4xy - 12y²
(2) Both Aexplanation for A. (V) x(3x+ 14) + 8
correct (vi) 5-4x(1 + 3x)
(3) Ais true, but
R is false (vii) x²y -3xy -40
(4) Ais false, but
R is true. (vii) (3x -2y) - 5(3x - 2y)- 24
Assertion (A) : 25x2-5x + 1is a perfect (ix) 12(a +b)² - (a + b) -35
(vi) Square trinomial. 6. Factorise :

Reason (R): Any trinomial which can be (0) 15(5x -4)- 10(5x 4)
expressed as x*+ y' + 2xy or x²+ y²-2xy (ü) 3a²x -bx +3a2-b
is a perfect square trinomial. (ii) b(c - d)' + a(d- c) +3(c - d)
(a) (1) (c) (3)
(b) (2) (d) (4) (iv) ax? +b²y- ab² -x2y
(vil)lAssertion (A) : x+7x + 12 (v) 1-3x-3y - 4(x + y
= x²+(4+3) x +4 x 3 7. Factorise:
=x² +4X + 3% + 4 x 3 (0) 2a- 50a (ii) 54a²b² -6
= (x +4) (x +3) (i) 64a?b- 144b3 (iv) (2x-y- (2x-y)
Reason (R):To factorise a given trinomial, (v) x²-2xy+y-2
the product of the first and the last term of (vi) x°-y-2yz -2
the trinomial is always the sum of the two 20x4
parts when we split the middle term. (vii) 7a5- 567a (viii) 5x2
(a) (1) (b) (2) (c) (3) (d) (4) 8. Factorise xy² - xz², Hence, find the value of :
(i) 9x 8 -9x 22
(x) Assertion (A) : The value of k so that the
(ii) 40 x 5.52- 40 x 4.52
121 11
factors of x-kx +16 the same is
2
9 Factorise
() (a-3b)-36 b²
Reason (R):(x+a) (x +b) =x*+(a+ b) x+ab (i) 25(a - 5b)- 4(a - 36)?
(a) (1) (b) (2) (c) (3) (d) (4) (ii) a'- 0-36 b²
(iv) x-5x2-36
() Assertion (A): There are two values ofb (v) 15(2x-y)²- 16(2x -y) -15
so that x²+ by - 24 is factorisable.
Reason (R):Two values have: 10. Evaluate (using factors) :3012 x 300 -3003.
11. Use factor method to evaluate
Product =-24 and sum = 2
la) (1) (b) (2) (c) (3) (0) (522- 80) + (z 4)
(d) (4)
2. Factorise: (52?-80) + (2-4)
0) 6x3 -8x? 5(2 -16) 5(2 -4²)
(0) 36x² -30x°y+ 48xy Z-4 Z-4
() 8(2a +3b)3 12(2a + 3b) 5(z +4) (2 -4) = 5(2 +4)
(w) 9a(x -2y) -12a (x -2y)° Z-4
3. Facorise :
0 a- ab (1-b) - b (i) 10y(6y + 21) + (2y + 7)
(0) xy' +(x-1)y 1 (ii) (a- 14a - 32) + (a +2)
() (ax +by)² +(bx (iv) 39x(50x- 98) + 26x² (5x + 7)
ay)
FactoDsation 177)