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).

Answered by Melissa P. Maths tutor

3689 Views

See similar Maths GCSE tutors

Related Maths GCSE answers

All answers ▸

Express 56 as the product of its prime factors


2a + 6b = 24, 3a + b = 20


How Would I Factorise A Quadratic Equation?


Make X the subject of the equation (9x-7)/3 = 8y


We're here to help

contact us iconContact usWhatsapp logoMessage us on Whatsapptelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

© MyTutorWeb Ltd 2013–2024

Terms & Conditions|Privacy Policy