Question

Factorise -:
i) 8(p-q)^3-12(p-q)^2
ii)1 \ 4a^2+a-3
with steps ofcourse

Answer 1

Answer:

i) 8(p-q)^3-12(p-q)^2

4(p-q)^2* [ 2(p-q)-3]

= 4(p-q)^2* ( 2p-2q-3)

ii)1\4a^2+a-3

= 1/4 [a^2+4a-12]

= 1/4 [ a^2+6a -2a -12]

=1/4 [a(a+6)-2(a+6)]

=1/4 [(a-2)(a+6)]