Question

5- (3a² – 2a) (6 – 3q² + 2a)
by factorise​

Answer 1

Answer:

5-(3a2-2a)(6-3a2+2a)

= 5 – (18a^2 – 9a^4 + 6a^3 – 12a + 6a^3 – 4a^2)

= 5 – (– 9a^4 + 6a^3 + 6a^3+ 18a^2 – 4a^2– 12a)

= 5 – (– 9a^4 + 12a^3 + 14a^2 – 12a)

= 9a^4 - 12a^3 - 14a^2 + 12a + 5

= (a+1)(3a+1)(a−1)(3a−5)