2.8 Factorisation of algebraic expressions:
We have learnt how to
factorise algebraic expressions similar to:
No 
Expression 
Factors 
1 
(pq)^{2} 3(pq) 
(pq){(pq)3} 
2 
2x(a4b)+3y(a4b) 
(a4b)(2x+3y) 
3 
m^{2}(pq+r)+mn(pq+r)+ n^{2}(pq+r) 
(pq+r) (m^{2}+mn+ n^{2}) 
Refer to section 2.5 to
recall how we factorised the expression of type px^{2}+mx +c.
2.8.1
Factorisation using identities/formulae:
We have learnt the following identities in section 2.3
No 
Formula/Identity 
Expansion 
Factors 
1 
(a+b)^{2} 
a^{2}+b^{2}+2ab 
(a+b)
and (a+b) 
2 
(ab)^{2} 
a^{2}+b^{2}2ab 
(ab)
and (ab) 
3 
(a+b)(ab) 
a^{2}b^{2} 
(a+b)
and (ab) 
4 
(x+a)*(x+b)

x^{2}+x(a+b)+ab 
(x+a)
and (x+b) 
2.8.1 Problem 1 : Factorize 9p^{2}+12pq +4q^{2}
using an identity
Solution:
The expression can be rewritten as 9p^{2}
+4q^{2}+12pq. This is of the form a^{2}+b^{2}+2ab with
a^{2}= 9p^{2} , b^{2}=
4q^{2} and 2ab=12pq
However 9p^{2 } can be split as 3p*3p =(3p)^{2} and 4q^{2}
= 2q*2q= (2q)^{2}and 12pq = 2*3p*2q
So we can say a=3p and b=2q
Since the given expression is of the form a^{2}+b^{2}+2ab
we can say the factors are (a+b) and (a+b)
_{}_{}The factors are (3p+2q) and (3p+2q)
Verification:
(3p+2q)(3p+2q)
=3p(3p+2q)+2q(3p+2q) ( Multiply each of the terms)
=9p^{2}+6pq +6qp+4q^{2} (simplification)
= 9p^{2}+12pq +4q^{2}
which is the term given in the
problem
2.8.1 Problem 2: Factorize 36x^{2}60x +25 using an identity
Solution:
The expression can be rewritten as 36x^{2}
+2560x. This is of the form a^{2}+b^{2}2ab with a^{2}=
36x^{2}, b^{2}= 25=5^{2}
and 2ab=60x
However 36x^{2 } can be split as 6x*6x = (6x)^{2} and
25 =5^{2} and 60x = 2*6x*5
So we can say a=6x and b=5.
Since the given expression is of the form a^{2}+b^{2}2ab
we can say the factors are (ab) and (ab).
_{}_{}The factors are (6x5) and (6x5).
Verification:
(6x5) (6x5)
=6x(6x5)5(6x5) ( Multiply each of the terms)
=36x^{2}30x
30x+25 (simplification)
= 36x^{2}60x +25 which is the term given
in the problem.
2.8.1 Problem 3 : Factorize (x+2)^{2}+18(x+2) +81 using an identity
Solution:
The expression can be rewritten as (x+2)^{2}
+81+18(x+2). This is of the form a^{2}+b^{2}+2ab with a^{2}=
(x+2)^{2} , b^{2}= 81=9^{2} and 2ab=18(x+2)
So we can say a=x+2 and b=9 and hence 2ab =
2(x+2)*9 =18(x+2)
Since the given expression is of the form a^{2}+b^{2}+2ab
we can say the factors are (a+b) and (a+b)
_{}_{}The factors are ((x+2)+9) and ((x+2)+9)
Verification:
Verify yourself that
((x+2)+9)((x+2)+9)= (x+2)^{2}+18(x+2) +81
2.8.1 Problem 4: Factorize p^{4}/16 q^{2}/64 using
an identity
Solution:
The expression
is of the form a^{2}b^{2} with a^{2}= p^{4}/16=
(p^{2}/4)^{2} and b^{2}=
q^{2}/64 = (q/8)^{2}
So we can say a=p^{2}/4 and b=q/8.
Since the given expression is of the form a^{2}b^{2}
we can say the factors are (a+b) and (ab).
_{}_{}The factors are (p^{2}/4+q/8)
and (p^{2}/4q/8).
Verification:
(p^{2}/4+q/8)(p^{2}/4q/8)
=p^{2}/4(p^{2}/4q/8)+q/8(p^{2}/4q/8) ( Multiply each of the terms)
=(p^{2}/4)^{2}p^{2}q/32
+qp^{2}/32 –(q/8)^{2}
(simplification)
= p^{4}/16 q^{2}/64 which is the term given in the problem
2.8.1 Problem 5: Factorize 8(x+1/x)^{2}18(x1/x)^{2} using
an identity
Solution:
We notice that 8 and 18 are not squares of any
number. But we observe 8 =2*4 and 18 =2*9. We also notice that 4 and 9 are
squares of 2 and 3 respectively.
Therefore we can rewrite the expression 8(x+1/x)^{2}18(x1/x)^{2}
= 2{4(x+1/x)^{2}9(x1/x)^{2}}.
The expression
4(x+1/x)^{2}9(x1/x)^{2} is of the form a^{2}b^{2}
with a^{2}= 4(x+1/x)^{2} =(2(x+1/x))^{2} ^{ }and
b^{2}=(3(x1/x))^{2}
So we can say a=2(x+1/x) and b=3(x1/x)
Since the given expression is of the form a^{2}b^{2}
we can say the factors are (a+b) and (ab)
_{}_{}The factors of 4(x+1/x)^{2}9(x1/x)^{2} are(2(x+1/x) + 3(x1/x)) and (2(x+1/x) 
3(x1/x))
Since we had taken 2 as common factor from the
given expression,
The factors of 8(x+1/x)^{2}18(x1/x)^{2} are 2
, (2(x+1/x) + 3(x1/x)) and (2(x+1/x)  3(x1/x))
Verification:
Exercise : Verify yourself that 2(2(x+1/x) +
3(x1/x))(2(x+1/x)  3(x1/x))= 8(x+1/x)^{2}18(x1/x)^{2}
2.8.1 Problem 5: If the difference of two numbers is 8 and the
difference between their squares is 400, find the numbers ( Lilavati Shloka 59)
Solution:
Let the numbers be x and y, then
x^{2}
y^{2}
=400
xy= 8 (_{} x=
y+8) (1)
x^{2}
y^{2}^{ }= (x+y)*(x –y) {a^{2}b^{2 }=(a+b)*(ab)}
= 8(x+y) (_{} xy =8)
_{} 400 = 8(x+y) (_{} x^{2} y^{2} =400)
_{} (x+y) = 50 (Division by 8 )
_{} y+8+y =50 (Substitute from (1)
_{} 2y = 42 (By simplification)
_{} y =21
_{} x= 29 ( Substitute in (1)
Verification:
2921 =8
29^{2}21^{2 }= ??
2.8.2 Product of three
binomials
We have learnt that
(x+a)*(x+b) = x^{2}+x(a+b)+ab
Using this identity let us find product of (x+a)*(x+b)*(x+c)`
(x+a)*(x+b)*(x+c)
= {(x+a)*(x+b)}*(x+c)
= {x^{2}+x(a+b)+ab}*(x+c)
= x^{2}(x+c)+x(a+b)*(x+c) + ab(x+c) ( every term of {x^{2}+x(a+b)+ab}
is multiplied with the every other term
of (x+c)
= x^{3}+ x^{2}c + x(a+b)*x+x(a+b)*c + abx+abc (
every term of x(a+b) is multiplied with the every other term of (x+c) )
= x^{3}+ x^{2}c + x^{2}(a+b)+x(a+b)*c
+ abx+abc (expansion)
= x^{3}+ x^{2}(c+a+b)+xac+xbc +
abx+abc (expansion and simplification)
= x^{3}+ x^{2}(a+b+c)+x(ac+
= x^{3}+ (a+b+c) x^{2}+(ab+bc+ca)x+abc
(Rearrangement)
Let us put b=a and c=a
in the identity (x+a)*(x+b)*(x+c)` then we get
(x+a)(x+a)(x+a)
= x^{3}+ (a+a+a) x^{2}+(a*a+a*a+a*a)x+a*a*a
= x^{3}+ 3ax^{2}+3a^{2}x+ a^{3}
= x^{3}+ 3ax(x+a)+ a^{3}
_{}(x+a)^{3} = x^{3}+ 3ax(x+a)+ a^{3}
If we replace x by a and a by b we get the
formula/identity
(a+b)^{3} = a^{3}+ 3ab(a+b)+ b^{3}
If we substitute –b for b in the above formula we
get
(ab)^{3}
= a^{3}+ 3a*b(ab)+ (b)^{3}
= a^{3}3ab(ab)b^{3}
2.8.2 Problem 1: Find the value of 1.05*0.97*.98
Solution:
1.05 = 1+.05, 0.97 = 10.03 and 0.98 = 10.02 and
they can be represented as x=1 and a=.05, b=0.03 and c= 0.02
Therefore the given product can be expressed as
(x+a)(x+b)(x+c)
We know (x+a)(x+b)(x+c)
= x^{3}+ (a+b+c) x^{2}+(ab+bc+ca)x+abc,
By substituting values for x, a, b , c we get
1.05*0.97*.98
= 1^{3}+ (0.050.030.02) 1^{2} +((0.05*0.03) (–0.03* 0.02)(0.02*0.05))1+ 0.05*0.03*0.02
= 1+ 0 1^{2}+(0.0015+0.00060.0010)1+
0.000030
= 1 0.0019+0.00003 =0.998130
Verification:
Verify using calculator
that 1.05*0.97*0.98 = 0.998130.
2.8.2 Problem 2 : Find the volume
of a cuboid whose sides are (5x+2)cms, (5x1)cms, (5x+3)cms
Solution:
We
know the volume of cuboid = length*breadth*depth. So
the given cuboid’s volume=(5x+2)(5x1)(5x+3)cc. The
product is of the form (x+a)(x+b)(x+c)
with x=5x and a=2, b=1 and c=3 Therefore
the given product can be expressed as (x+a)(x+b)(x+c) We
know (x+a)(x+b)(x+c)= x^{3}+
(a+b+c) x^{2}+(ab+bc+ca)x+abc, =
(5x)^{3}+ (21+3) (5x)^{2}+(23+6)(5x)+ 2*1*3(By
substituting values for x, a, b , c) =
125x^{3}+ 100x^{2}+5x6 

Verification: (for one value of x)
Let the value of x in the sides of cuboid be 2
Hence the sides are:
1.5x+2=5*2+2=12.
2.5x1 =5*21=9
3.5x+3 =5*2+3= 13
From the solution arrived above the volume of the
cuboid
= 125x^{3}+ 100x^{2}+5x 6 =
125*8+100*4+5*26
= 1000+400+106=1404 (As arrived from the formula)
However, we know the volume of cuboid whose sides
are12,9 and 13 is
=12*9*13 = 1404 cubic units.
Since the volume arrived from two different methods
are same our solution is correct.
We have seen that (x+a)(x+b)(x+c)= x^{3}+
(a+b+c) x^{2}+(ab+bc+ca)x+abc and we observe
1. The coefficient of x^{2} in (a+b+c) x^{2}
is (a+b+c)
2. The coefficient of x in (ab+bc+ca)x is
(ab+bc+ca)
2.8.2 Problem 3: Find the coefficient of x^{2} and x in (3x1)(3x1)(3x+4)
Solution:
The product is of the form (x+a)(x+b)(x+c) with x=3x and a=1, b=1 and c=4
Therefore the given product can be expressed as
(x+a)(x+b)(x+c)
We know (x+a)(x+b)(x+c)= x^{3}+ (a+b+c) x^{2}+(ab+bc+ca)x+abc.
= (3x)^{3}+(a+b+c)(3x)^{2} +
(ab+bc+ca)(3x)+abc (By substituting 3x for x)
1. The coefficient of x^{2} in (a+b+c) (3x)^{2}
is (a+b+c)*9. By substituting values for a,b and c we get:
The coefficient of x^{2} is (a+b+c)*9 =
(11+4)*9 = 18
2. The coefficient of x in (ab+bc+ca)(3x) is
(ab+bc+ca)*3. By substituting values for a,b and c we get:
The coefficient of x in (ab+bc+ca)3x is
(ab+bc+ca)*3 = (144)*3 = 21
Verification:
Expand
(3x1)(3x1)(3x+4) and check the coefficients
We have seen earlier that
(a+b)^{3} = a^{3}+ 3ab(a+b)+ b^{3}
_{} (a+b)^{3}
3ab(a+b) = a^{3}+ b^{3}(By transposition)
i,e a^{3}+ b^{3}
=(a+b)^{3} 3ab(a+b)
= (a+b){ (a+b)^{2}
3ab}
= (a+b) { a^{2}
+b^{2} +2ab 3ab}(By expanding (a+b)^{2})
= (a+b) (a^{2} +b^{2} ab)
By Substituting –b for b in the above identity we
get
a^{3}+ (b)^{3}
= (a+b) (a^{2} +(b)^{2} a*(b))
= (ab) (a^{2} +b^{2} +ab)
But a^{3}+ (b)^{3}= a^{3}b^{3}
_{} a^{3}b^{3}= (ab) (a^{2} +b^{2}
+ab)
2.8.2 Problem 3: Factorise 0.027 p^{3}+0.008 q^{3}
Solution:
Note 0.3*0.3*0.3=0.027 and
0.2*0.2*0.2=0.008
Therefore the given expression is of the form a^{3}+b^{3
}with a=0.3p and b= 0.2q
By using a^{3}+b^{3}=(a+b) (a^{2}
+b^{2} ab) and substituting for
a and b we get
0.027 p^{3}+0.008 q^{3}
= (0.3p+0.2q) ((0.3p)^{2} +(0.2q)^{2}
0.3p*0.2q)
= (0.3p+0.2q) (0.09p^{2} +0.04q^{2}
0.06pq)
Verification: (for one value of p and q)
Let p=1 and q=1, By substituting these in the solution we get
(0.3p+0.2q) (0.09p^{2} +0.04q^{2}
0.06pq)
= 0.5*(0.09+0.040.06) = 0.5*0.07 = 0.035
Also
0.027 p^{3}+0.008 q^{3}
=0.027+0.008 =0.035
Since the results got by both methods are same, our
solution is correct
2.8.2 Problem 4: Factorise 125 1/ a^{3}b^{3}
Solution:
Note that 125 = 5^{3}
and 1/ a^{3}b^{3}=(1/ ab)^{3}
Therefore the given expression is of the form a^{3}b^{3
}with a=5 and b= 1/ab
By using a^{3}b^{3}=(ab) (a^{2}
+b^{2} +ab) and substituting for
a and b we get
125 1/ a^{3}b^{3}
= (5 1/ab) (5^{2} +(1/ab)^{2}
+5*1/ab)
= (5 1/ab) (25 +1/a^{2} b^{2}
+5/ab)
Verification: (for one value of a and b)
Let a=1 and b=2, By substituting these in the
solution
(5 1/ab) (25 +1/a^{2} b^{2} +5/ab)
=(51/2)(25+1/4+5/2) =124.875(Use calculator)
Also
125 1/ a^{3}b^{3}
= 1251/8= 124.875(Use calculator)
Since the results got by both methods are same our
solution is correct
2.8 Summary of
learning
No 
Formula/Identity 
Expansion 
Factors 
1 
(a+b)^{2} 
a^{2}+b^{2}+2ab 
(a+b)
and (a+b) 
2 
(ab)^{2} 
a^{2}+b^{2}2ab 
(ab)
and (ab) 
3 
(a+b)(ab) 
a^{2}b^{2} 
(a+b)
and (ab) 
4 
(x+a)*(x+b)

x^{2}+x(a+b)+ab 
(x+a)
and (x+b) 
5 
(x+a)(x+b)(x+c) 
x^{3}+
(a+b+c)x^{2}+(ab+bc+ca)x+abc 
(x+a),(x+b)and
(x+c) 
6 
(a+b)^{3} 
a^{3}+b^{3}+3ab(a+b) 
(a+b),(a+b)
and(a+b) 
7 
(ab)^{3} 
a^{3}b^{3}3ab(ab) 
(ab),(ab)
and(ab) 
8 
a^{3}+b^{3} 
(a+b)
(a^{2} +b^{2} ab) 
(a+b)
and (a^{2} +b^{2} ab) 
9 
a^{3}b^{3} 
(ab) (a^{2} +b^{2} +ab) 
(ab)
and (a^{2} +b^{2} +ab) 