1. ## Factorising an expression?

Sometimes I think that just a little discussion can make all the difference if like minded people can help each other out

I am asked to factorise the following expression;

6p^3 - 4pq + 2p. This I understand to be reverse engineering, but I seem to struggle with it at the moment?

6p^3 - 4pq + 2p = 6p^2 (p - 2pq + 2)

Now if I say; 6p^2 x p = 6p^3 - 2 x 2pq +2)

result; 6p^3 - 4pq + 2

My problem seems to be what happend to the "p" at the end with 2? i.e. 2p?

Thanks for any help

I think I have worked out the following solution?

6p^3 - 2q + 1 = 6p^2 (p - 2q + 1) = 6p^3 - 4pq +2p

Not sure if i am right though?

2. Originally Posted by David Green
Sometimes I think that just a little discussion can make all the difference if like minded people can help each other out

I am asked to factorise the following expression;

6p^3 - 4pq + 2p. This I understand to be reverse engineering, but I seem to struggle with it at the moment?

6p^3 - 4pq + 2p = 6p^2 (p - 2pq + 2)

Now if I say; 6p^2 x p = 6p^3 - 2 x 2pq +2)

result; 6p^3 - 4pq + 2

My problem seems to be what happend to the "p" at the end with 2? i.e. 2p?

Thanks for any help
Hi David Green,

$6p^3-4pq+2p$

First, you need to factor out the common monomial factor which is $2p$, not $6p^2$

Then you will arrive at $2p(3p^2-2q+1)$

And from there I don't see anything else to do.

3. Originally Posted by masters
Hi David Green,

$6p^3-4pq+2p$

First, you need to factor out the common monomial factor which is $2p$, not $6p^2$

Then you will arrive at $2p(3p^2-2q+1)$

And from there I don't see anything else to do.
Thanks for that, it is just what I required. Now I understand I think?

6p^3 - 4pq + 2p reverse engineered

move the 2p to the LHS; 2p (3p^2, this equals 6p^3, then 2p x 2q = 4pq and the + 1 is the remaining p

Thanks alot much appreciated.

David