Example 6. Resolve into factors : x^2 5 x 6.
Solution : x^2 5 x 6 x^2 ( 2 3 )x( 2 )( 3 )
= (x 2 )(x 3 )
Example 7. Resolve into factors : x^2 2 x 35.
Solution : x^2 2 x 35
= x^2 ( 7 5 )x( 7 )( 5 )
= (x 7 )(x 5 )
Example 8. Resolve into factors : x^2 x 20.
Solution : x^2 x 20
= x^2 ( 5 4 )x( 5 )( 4 )
= (x 5 )(x 4 )
(e) By middle term break-up method of polynomial of the form of ax^2 bxc:
If ax^2 bxc (rxp)(sxq)
ax^2 bxc rsx^2 (rqsp)xpqThat is, a rs,b rqsp and c pq.
Hence,ac rspq (rq)(sp) and b rqsp
Therefore, to determine factors of the polynomial ax^2 bxc, ac that is, the
product of the coefficient of x^2 and the term free from x are to be expressed into two
such factors whose algebraic sum is equal to b, the coefficient of x.
Example 9. Resolve into factors : 12 x^2 35 x 18.
Solution : 12 x^2 35 x 18
Here, 12 u 18 216 27 u 8 and 27 8 35
? 12 x^2 35 x 18 12 x^2 27 x 8 x 18
= 3 x( 4 x 9 ) 2 ( 4 x 9 )
= ( 4 x 9 )( 3 x 2 )
Example 10. Resolve into factors : 3 x^2 x 14.
Solution : 3 x^2 x 14 3 x^2 7 x 6 x 14
= x( 3 x 7 ) 2 ( 3 x 7 )
= ( 3 x 7 )(x 2 )
Activity : Resolve into factors :
1.x^2 x 56 2. 16 x^3 46 x^2 15 x 3. 12 x^2 17 x 6
(f) Expressing the expression in the form of perfect cubes :