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# Factoring watch

1. I've simplified the question down to this:

p(p+5)-q(q+5)

What are the steps involved to get from where I am to the answer? Can p(p+5)-q(q+5) be rewritten as (p-q)(p+5)(q+5)?
2. (Original post by damo215)
I've simplified the question down to this:

p(p+5)-q(q+5)

What are the steps involved to get from where I am to the answer? Can p(p+5)-q(q+5) be rewritten as (p-q)(p+5)(q+5)?
Yes because you are multiplying the all the terms together. Remember, x(x+5) is basically x multiplied by x+5 which is x^2 + 5x.
Similarly, that equation is p * (p+5) * -q * (q+5).
This mat become simpler when you learn how to factorise quadratic trinomial equations using the grouping method. It's so similar, however the two brackets will be identical and the two separate terms will be combined to form a bracket.

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3. I've expanded this: p(p+5)-q(q+5)

to get this: p^2+5p-q^2-5q - expansion 1

I've expanded the answer and I get this: p^2+pq+5p-qp-q^2-5q - expansion 2

The negative and positive pq's cancel each other out leaving me with the same equation as expansion 1.

How do they factor expansion 2 (and expansion 1) to get the answer, specifically the trinomial factor (p+q+5)

Sorry if this is going around in circles.

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Updated: January 20, 2016
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