2.8 Factorisation of algebraic expressions:
We have learnt how to
factorise algebraic expressions similar to:
|
No |
Expression |
Factors |
|
1 |
(p-q)2- 3(p-q) |
(p-q){(p-q)-3} |
|
2 |
2x(a-4b)+3y(a-4b) |
(a-4b)(2x+3y) |
|
3 |
m2(pq+r)+mn(pq+r)+ n2(pq+r) |
(pq+r) (m2+mn+ n2) |
Refer to section 2.5 to
recall how we factorised the expression of type px2+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 |
a2+b2+2ab |
(a+b)
and (a+b) |
|
2 |
(a-b)2 |
a2+b2-2ab |
(a-b)
and (a-b) |
|
3 |
(a+b)(a-b) |
a2-b2 |
(a+b)
and (a-b) |
|
4 |
(x+a)*(x+b)
|
x2+x(a+b)+ab |
(x+a)
and (x+b) |
2.8.1 Problem 1 : Factorize 9p2+12pq +4q2
using an identity
Solution:
The expression can be rewritten as 9p2
+4q2+12pq. This is of the form a2+b2+2ab with
a2= 9p2 , b2=
4q2 and 2ab=12pq
However 9p2 can be split as 3p*3p =(3p)2 and 4q2
= 2q*2q= (2q)2and 12pq = 2*3p*2q
So we can say a=3p and b=2q
Since the given expression is of the form a2+b2+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)
=9p2+6pq +6qp+4q2 (simplification)
= 9p2+12pq +4q2
which is the term given in the
problem
2.8.1 Problem 2: Factorize 36x2-60x +25 using an identity
Solution:
The expression can be rewritten as 36x2
+25-60x. This is of the form a2+b2-2ab with a2=
36x2, b2= 25=52
and -2ab=-60x
However 36x2 can be split as 6x*6x = (6x)2 and
25 =52 and -60x = -2*6x*5
So we can say a=6x and b=5.
Since the given expression is of the form a2+b2-2ab
we can say the factors are (a-b) and (a-b).
The factors are (6x-5) and (6x-5).
Verification:
(6x-5) (6x-5)
=6x(6x-5)-5(6x-5) ( Multiply each of the terms)
=36x2-30x
-30x+25 (simplification)
= 36x2-60x +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 a2+b2+2ab with a2=
(x+2)2 , b2= 81=92 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 a2+b2+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 p4/16- q2/64 using
an identity
Solution:
The expression
is of the form a2-b2 with a2= p4/16=
(p2/4)2 and b2=
q2/64 = (q/8)2
So we can say a=p2/4 and b=q/8.
Since the given expression is of the form a2-b2
we can say the factors are (a+b) and (a-b).
The factors are (p2/4+q/8)
and (p2/4-q/8).
Verification:
(p2/4+q/8)(p2/4-q/8)
=p2/4(p2/4-q/8)+q/8(p2/4-q/8) ( Multiply each of the terms)
=(p2/4)2-p2q/32
+qp2/32 –(q/8)2
(simplification)
= p4/16- q2/64 which is the term given in the problem
2.8.1 Problem 5: Factorize 8(x+1/x)2-18(x-1/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(x-1/x)2
= 2{4(x+1/x)2-9(x-1/x)2}.
The expression
4(x+1/x)2-9(x-1/x)2 is of the form a2-b2
with a2= 4(x+1/x)2 =(2(x+1/x))2 and
b2=(3(x-1/x))2
So we can say a=2(x+1/x) and b=3(x-1/x)
Since the given expression is of the form a2-b2
we can say the factors are (a+b) and (a-b)
The factors of 4(x+1/x)2-9(x-1/x)2 are(2(x+1/x) + 3(x-1/x)) and (2(x+1/x) -
3(x-1/x))
Since we had taken 2 as common factor from the
given expression,
The factors of 8(x+1/x)2-18(x-1/x)2 are 2
, (2(x+1/x) + 3(x-1/x)) and (2(x+1/x) - 3(x-1/x))
Verification:
Exercise : Verify yourself that 2(2(x+1/x) +
3(x-1/x))(2(x+1/x) - 3(x-1/x))= 8(x+1/x)2-18(x-1/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
x2
-y2
=400
x-y= 8 ( x=
y+8) ------(1)
x2
-y2 = (x+y)*(x –y) {a2-b2 =(a+b)*(a-b)}
= 8(x+y) (
x-y =8)
400 = 8(x+y) ( x2 -y2 =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:
29-21 =8
292-212 = ??
2.8.2 Product of three
binomials
We have learnt that
(x+a)*(x+b) = x2+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)
= {x2+x(a+b)+ab}*(x+c)
= x2(x+c)+x(a+b)*(x+c) + ab(x+c) ( every term of {x2+x(a+b)+ab}
is multiplied with the every other term
of (x+c)
= x3+ x2c + 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) )
= x3+ x2c + x2(a+b)+x(a+b)*c
+ abx+abc (expansion)
= x3+ x2(c+a+b)+xac+xbc +
abx+abc (expansion and simplification)
= x3+ x2(a+b+c)+x(ac+
= x3+ (a+b+c) x2+(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)
= x3+ (a+a+a) x2+(a*a+a*a+a*a)x+a*a*a
= x3+ 3ax2+3a2x+ a3
= x3+ 3ax(x+a)+ a3
(x+a)3 = x3+ 3ax(x+a)+ a3
If we replace x by a and a by b we get the
formula/identity
(a+b)3 = a3+ 3ab(a+b)+ b3
If we substitute –b for b in the above formula we
get
(a-b)3
= a3+ 3a*-b(a-b)+ (-b)3
= a3-3ab(a-b)-b3
2.8.2 Problem 1: Find the value of 1.05*0.97*.98
Solution:
1.05 = 1+.05, 0.97 = 1-0.03 and 0.98 = 1-0.02 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)
= x3+ (a+b+c) x2+(ab+bc+ca)x+abc,
By substituting values for x, a, b , c we get
1.05*0.97*.98
= 13+ (0.05-0.03-0.02) 12 +((0.05*-0.03) (–0.03* -0.02)(-0.02*0.05))1+ 0.05*-0.03*-0.02
= 1+ 0 12+(-0.0015+0.0006-0.0010)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, (5x-1)cms, (5x+3)cms
Solution:
|
We
know the volume of cuboid = length*breadth*depth. So
the given cuboid’s volume=(5x+2)(5x-1)(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)= x3+
(a+b+c) x2+(ab+bc+ca)x+abc, =
(5x)3+ (2-1+3) (5x)2+(-2-3+6)(5x)+ 2*-1*3(By
substituting values for x, a, b , c) =
125x3+ 100x2+5x-6 |
|
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.
2.5x-1 =5*2-1=9
3.5x+3 =5*2+3= 13
From the solution arrived above the volume of the
cuboid
= 125x3+ 100x2+5x- 6 =
125*8+100*4+5*2-6
= 1000+400+10-6=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)= x3+
(a+b+c) x2+(ab+bc+ca)x+abc and we observe
1. The co-efficient of x2 in (a+b+c) x2
is (a+b+c)
2. The co-efficient of x in (ab+bc+ca)x is
(ab+bc+ca)
2.8.2 Problem 3: Find the co-efficient of x2 and x in (3x-1)(3x-1)(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)= x3+ (a+b+c) x2+(ab+bc+ca)x+abc.
= (3x)3+(a+b+c)(3x)2 +
(ab+bc+ca)(3x)+abc (By substituting 3x for x)
1. The co-efficient of x2 in (a+b+c) (3x)2
is (a+b+c)*9. By substituting values for a,b and c we get:
The co-efficient of x2 is (a+b+c)*9 =
(-1-1+4)*9 = 18
2. The co-efficient of x in (ab+bc+ca)(3x) is
(ab+bc+ca)*3. By substituting values for a,b and c we get:
The co-efficient of x in (ab+bc+ca)3x is
(ab+bc+ca)*3 = (1-4-4)*3 = -21
Verification:
Expand
(3x-1)(3x-1)(3x+4) and check the co-efficients
We have seen earlier that
(a+b)3 = a3+ 3ab(a+b)+ b3
(a+b)3
-3ab(a+b) = a3+ b3(By transposition)
i,e a3+ b3
=(a+b)3 -3ab(a+b)
= (a+b){ (a+b)2
-3ab}
= (a+b) { a2
+b2 +2ab -3ab}(By expanding (a+b)2)
= (a+b) (a2 +b2 -ab)
By Substituting –b for b in the above identity we
get
a3+ (-b)3
= (a+-b) (a2 +(-b)2 -a*(-b))
= (a-b) (a2 +b2 +ab)
But a3+ (-b)3= a3-b3
a3-b3= (a-b) (a2 +b2
+ab)
2.8.2 Problem 3: Factorise 0.027 p3+0.008 q3
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 a3+b3
with a=0.3p and b= 0.2q
By using a3+b3=(a+b) (a2
+b2 -ab) and substituting for
a and b we get
0.027 p3+0.008 q3
= (0.3p+0.2q) ((0.3p)2 +(0.2q)2
-0.3p*0.2q)
= (0.3p+0.2q) (0.09p2 +0.04q2
-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) (0.09p2 +0.04q2
-0.06pq)
= 0.5*(0.09+0.04-0.06) = 0.5*0.07 = 0.035
Also
0.027 p3+0.008 q3
=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/ a3b3
Solution:
Note that 125 = 53
and 1/ a3b3=(1/ ab)3
Therefore the given expression is of the form a3-b3
with a=5 and b= 1/ab
By using a3-b3=(a-b) (a2
+b2 +ab) and substituting for
a and b we get
125 -1/ a3b3
= (5 -1/ab) (52 +(1/ab)2
+5*1/ab)
= (5 -1/ab) (25 +1/a2 b2
+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/a2 b2 +5/ab)
=(5-1/2)(25+1/4+5/2) =124.875(Use calculator)
Also
125 -1/ a3b3
= 125-1/8= 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 |
a2+b2+2ab |
(a+b)
and (a+b) |
|
2 |
(a-b)2 |
a2+b2-2ab |
(a-b)
and (a-b) |
|
3 |
(a+b)(a-b) |
a2-b2 |
(a+b)
and (a-b) |
|
4 |
(x+a)*(x+b)
|
x2+x(a+b)+ab |
(x+a)
and (x+b) |
|
5 |
(x+a)(x+b)(x+c) |
x3+
(a+b+c)x2+(ab+bc+ca)x+abc |
(x+a),(x+b)and
(x+c) |
|
6 |
(a+b)3 |
a3+b3+3ab(a+b) |
(a+b),(a+b)
and(a+b) |
|
7 |
(a-b)3 |
a3-b3-3ab(a-b) |
(a-b),(a-b)
and(a-b) |
|
8 |
a3+b3 |
(a+b)
(a2 +b2 -ab) |
(a+b)
and (a2 +b2 -ab) |
|
9 |
a3-b3 |
(a-b) (a2 +b2 +ab) |
(a-b)
and (a2 +b2 +ab) |
