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 Post subject: FactorisingPosted: Thu Dec 06, 2018 6:26 pm

Joined: Thu Sep 11, 2014 2:03 pm
Posts: 457
Please can someone explain to me (in simple steps) how to Factorise the following:

pq+6−2q−3p

I realise this is beyond 11+ Maths prep, I've googled it and am now totally confused!

Many Thanks

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 Post subject: Re: FactorisingPosted: Thu Dec 06, 2018 8:09 pm

Joined: Fri Aug 24, 2018 5:26 pm
Posts: 125
(pq-2q)+(6-3p)=q(p-2)+3(2-p)=(p-2)(q-3)

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 Post subject: Re: FactorisingPosted: Thu Dec 06, 2018 10:37 pm

Joined: Thu Sep 11, 2014 2:03 pm
Posts: 457
Yes wrote:
(pq-2q)+(6-3p)=q(p-2)+3(2-p)=(p-2)(q-3)

Thank you, I don't have a very good maths brain!
I'll see if I can work out how you got there in the morning.

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 Post subject: Re: FactorisingPosted: Fri Dec 07, 2018 11:09 am

Joined: Sat Aug 10, 2013 12:46 pm
Posts: 243
pq+6−2q−3p has no common factors, but bits of it do. You could look at the p terms.
pq-3p-2q+6=p(q-3)-2q+6

This leaves 6-2q, which has a common factor of 2.
p(q-3)-2q+6=p(q-3)+2(-q+3)

2(-q+3) is the same as -2(q-3). Think of it as taking an extra factor of -1 outside the bracket.
p(q-3)+2(-q+3)=p(q-3)-2(q-3)

So now there is a common factor of (q-3). (If you think of (q-3) like x, factorising px-2x is easy)

p(q-3)-2(q-3)=(p-2)(q-3)

You could have also done it by factorising the q terms first. Try it and see.

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 Post subject: Re: FactorisingPosted: Fri Dec 07, 2018 12:57 pm

Joined: Fri Mar 10, 2017 5:06 pm
Posts: 682
I’ll try to explain it in what is hopefully a more simple way.

Firstly remember that factorising is essentially the reverse of multiplying out brackets.

So in general terms:

(a+ b)(c + d) = ac + ad + bc + bd

Our equation is pq + 6 – 2q -3p.

The fact that we have pq as one of our terms means that when we factorise we must have p in one of the brackets and q in the other, in order for it to multiply out to give pq as one of the terms.

So this is going to give us something like:

(p + b)(q + d) = pq + 6 – 2q -3p

Where b and d are unknown numbers.

Rearranging the right hand side slightly gives:

(p + b)(q + d) = pq -3p – 2q + 6

It should now be straightforward to see that in order to get the right numbers of p’s and q’s ,d = -3 and b = -2.
Also this would give b * d = -2 * -3 = +6, which is thankfully the final term in our equation, so proves we’re right.

So this would give a final solution of

(p -2)(q – 3).

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 Post subject: Re: FactorisingPosted: Fri Dec 07, 2018 9:49 pm

Joined: Thu Sep 11, 2014 2:03 pm
Posts: 457
Wow, Thank you so much.

I actually understand it now!

I'm going to try a few more and see if i've mastered it.

Thanks again

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