Thread: More factorising with common factor

Thank you so much for your help with that one, I have one more if you don't mind, having trouble with this as well.

(2p-1)(p+3) + (7p+4)(p+3)

I know they both have the p + 3 in common but I don't know where to start. Thank you very much for your help.

Originally Posted by MathMack

(2p-1)(p+3) + (7p+4)(p+3)

First try to multiply it yourself then see the thread, you will get it
(p+3)*[(2p - 1) + (7p + 4)]

Thanks for the reply, I did do that and I got that, but the book has the answer as 3(p+3)(3p+1), not sure how to get there. Thanks

Originally Posted by MathMack

Thank you so much for your help with that one, I have one more if you don't mind, having trouble with this as well.

(2p-1)(p+3) + (7p+4)(p+3)

I know they both have the p + 3 in common but I don't know where to start. Thank you very much for your help.

(p + 3) ([2p - 1] + [7p + 4]) = (p + 3)(2p - 1 + 7p + 4)

= (p + 3)(9p + 3).

Since 3 is a commn factor or 9p + 3 you can write it as 3(3p + 1).

So you have (p + 3) 3(3p + 1) = 3(p + 3)(3p + 1)

since obviously you can change the order of things that you multiply together.