Factorise fully 3*a^3*b +12*a^2*b^2 + 9*a^5*b^3

To factorise 3a3b + 12a2b2 + 9a5b3, we need to deal with like elements together.Start with the integers. The highest common factor of 3, 12, and 9 is 3. Therefore we factor out the 3 and the expression becomes3(a3b + 4a2b2 + 3a5b3)Next, deal with the a values. The highest common factor of a3, a2 and a5 is a2. So, we factor out a2and the expression becomes3a2(ab + 4b2 + 3a3b3)Finally, we need to factorise the b values. The highest common factor of b, b2 and bis b. So we factor out b and the expression becomes.3a2b(a + 4b + 3a3b2). Therefore, the complete factorisation of 3a3b + 12a2b2 + 9a5bis 3a2b(a + 4b + 3a3b2).

MP
Answered by Melissa P. Maths tutor

4949 Views

See similar Maths GCSE tutors

Related Maths GCSE answers

All answers ▸

Factorise the following equation: y = 2x^2 + 4x - 6


Ayo is 7 years older than Hugo. Mel is twice as old as Ayo. The sum of their three ages is 77 Find the ratio of Hugo's age to Ayo's age to Mel's age.


Find ∫(4x - 2/(x^3)) dx


√(6^2+8^2)=^3√125a^3


We're here to help

contact us iconContact ustelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

MyTutor is part of the IXL family of brands:

© 2026 by IXL Learning