### Factorising expressions.

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Messages: 1 - 2 of 2
• #### Message 1.

Posted by U15276000 (U15276000) on Sunday, 20th May 2012

I'm currently revising for upcoming exams next week and I have a question that looks like this...
14) a) 5x² - 5xy
b) 8ab² + 2
c) 9y³ - 12x²y
d) 6p²q + 10pq²

If anyone can actually explain how to do these as well as provide an answer, I'd be really grateful. Thank you ♥.

• #### Message 2

, in reply to message 1.

Posted by U15280596 (U15280596) on Thursday, 24th May 2012

Well, when you factorise a number, you separate out the things that multiply together to make that number. It sort of works similarly in algebra. You have:

a) 5x² - 5xy, and both bits have 5x in so 5x(x - y) is probably the factorisation they want. You could make it 5.x.(x - y) which would not be wrong, but would be unsual.

b) 8ab² + 2, now 2 divides both bits, so you have 2(4ab² + 1) and you can't take any more factors out of the bits in the bracket.

c) 9y³ - 12x²y, now you have a y in both bits and 9=3x3 and 12=4x3, so you have a 3 in both bits too, taking those both out gives 3y outside... 3y(3y² - 4x²)

d) 6p²q + 10pq², now 6=2x3, 10=2x5, so you have a 2 in both bits. you also have a p in both bits and a q in both bits. Taking those all out, you have 2pq(3p+5q)

hope this helps

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