Page 58 - SM inner class 8.cdr

A number of expressions to be factorized are (ii) Factorization using Identity II :
of the form or can be put into the form : In the 2 nd identity, a - 2 ab + b = ( a - b) 1st
2
2
2
2
2
2
2
2
,
,
a + 2 ab + b a - 2 ab + b a 2 - b and and the last term should be perfect square and
2
x + ( a + b + ab. These expressions can be
)
the middle term is two times the square root of
easily factorized using Identities I, II, III and st
1 and the last term and the sign of the middle
IV given below.
term is negative.
2
2
(i) a + 2 ab + b = ( a + b) 2
2
2
-
(ii) a - 2 ab + b ( a b) 2 Examples on 2nd Identity of
Factorization :
2
2
(
-
)
(iii) a - b = ( a + b a b) 2 2
2
)
(
)
(iv) x + ( a + b x + ab = ( x + a x + b) Example 9. 4 p - 20pq + 25q
2
Solution: 4 p - 20pq + 25q 2
In this section we will learn Factorization
2
= (2p ) 2 - .(2p ).(5q ) + (5q ) 2
using Identities one by one.
2
)
= (2p - 5q [ since a = 2 p and b = 5 q;
(i) Factorization using Identity I : Q = (a - ) ]
2
b
2
2
2
st
In the 1 identity, a + 2 ab + b = ( a + b) , We Example 10. 1 16- x 2 + 64x 4
st
can see that the 1 and the last term should be Solution: = ( )1 2 - .( ).(8x 2 ) + (8x 2 2
1
)
2
perfect square and the middle term is two = (1 8x 2 2 x ;
2
) [ since a =1 and b = 8
-
st
times the square root of the product of 1 and 2 2 2
a - 2 ab + b = ( a - b) ]
the last term and the sign of the middle term is
positive. (iii) Factorization using Identity III :
Examples on 1st Identity of Some quadratic polynomials will be missing
the middle term. Often these polynomials are
Factorization
the difference of two squares.
2
Example 7. 9a + 12ab + 4b 2
These polynomials come from multiplying the
2
Solution: 9a + 12ab + 4b 2
sum and difference of binomials, such as
2
= (3a ) 2 + .(3a ).(2b ) + (2b ) 2 (a + b ) (a - ) b = a - b 2 when simplified.
2
Now let, 3a = A and 2b = B
So, substituting A for 3a and B for 2b we get: Examples on 3rd Identity of
2
2
A + 2 AB + B Now, this is similar to the I Factorization :
Identity i.e. Example 11. 16x - 9y 2
2
2
2
a + 2 ab + b = ( a + b) 2 Solution: 16x - 9y 2
2
2
So, factor is (A + B ) = ( a + 2 ) b 2 = (4x ) 2 - (3y ) 2
3
or (3a + 2b ) (3a + 2b ) = (4x + 3y ) (4x - 3y )
[ since A = 3 a and B = 2 b] [since a = 4 x and b = 3 y;
2
2
1 a - b = ( a + b a - b)]
)
(
4
Example 8. x + 2 +
x 4 Example 12. x - x y 4
4
4
1 4 4 4
4
Solution: x + 2 + Solution : x - x y
x 4 = (x 2 2 -[(xy ) ]
2 2
)
2
1 æ 1 ö 2 2 2 2 2 2
)
2
= (x 2 2 + .(x 2 ) . + ç ÷ = (x + x y ) (x - x y )
x 2 è x 2 ø [ since a = x and b = ( xy) ;
2
2
2
2
2
)
(
= (x 2 + / x 2 2 x and b =1/ x] a - b = ( a + b a - b)]
1
) [since a =
æ 1 ö æ 1 ö In the 2nd parenthesis(bracket), we can
= çx 2 + ÷ çx 2 + ÷
è x 2 ø è x 2 ø apply the 3rd identity of Factorization again
2
= (x 2 + x y 2 ) (x + xy ) (x - xy )
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Mathematics In Focus - 8