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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 becomes

3(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 becomes

3a2(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).

Melissa P. GCSE Maths tutor, 13 plus  Maths tutor, 11 Plus Maths tuto...

10 months ago

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