(ii) 2 ab(xy) 2 bc(xy) 3 ca(xy) (xy)( 2 ab 2 bc 3 ca)(b) Expressing an expression in the form of a perfect square ;
Example 1. Resolve into factors : 4 x^2 12 x 9.
Solution : 4 x^2 12 x 9 ( 2 x)^2 2 u 2 xu 3 ( 3 )^2
= ( 2 x 3 )^2 ( 2 x 3 )( 2 x 3 )
Example 2. Resolve into factors : 9 x^2 30 xy 25 y^2.
Solution : 9 x^2 30 xy 25 y^2
= ( 3 x)^2 2 u 3 xu 5 y( 5 y)^2
= ( 3 x 5 y)^2 ( 3 x 5 y)( 3 x 5 y)
(c) Expressing an expression as the difference of two squares and then applying
the formula a^2 b^2 (ab)(ab):
Example 3. Reslove into factors : a^2 1 2 bb^2.
Solution : a^2 1 2 bb^2 a^2 (b^2 2 b 1 )
= a^2 (b 1 )^2 {a(b 1 )}{a(b 1 )}
= (ab 1 )(ab 1 )
Example 4. Resolve into factors : a^4 64 b^4.
Solution : a^4 64 b^4 (a^2 )^2 ( 8 b^2 )^2
= (a^2 )^2 2 ua^2 u 8 b^2 ( 8 b^2 )^2 16 a^2 b^2
= (a^2 8 b^2 )^2 ( 4 ab)^2
= (a^2 8 b^2 4 ab)(a^2 8 b^2 4 ab)
= (a^2 4 ab 8 b^2 )(a^2 4 ab 8 b^2 )
Activity : Resolve into factors :
1.abx^2 acx^3 adx^4 2. xa^2 144 xb^2 3. x^2 2 xy 4 y 4(d) Using the formula x^2 (ab)xab (xa)(xb):
Example 5. Resolve into factors : x^2 12 x 35.
Solution : x^2 12 x 35 x^2 ( 5 7 )x 5 u 7
= (x 5 )(x 7 )
In this method, a polynomial of the form x^2 pxq can be factorized, if two
integers a and b can be found so that, it is ab p and ab q. For this, two
factors of q with their signs are to be taken whose algebraic sum is p. If q! 0 ,a
andb will be of same signs and if q 0 ,a and b will be of opposite signs.