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) 