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Can you explain the grouping method of factoring. Can you describe a scenario when the grouping method would be preferred over other methods and show me an equation in this situation?
I need help with this, if you can please help(:
 Feb 6, 2014
 #1
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http://www.jamesbrennan.org/algebra/polynomials/factoring_by_grouping_.htm

has a pretty good explanation of how to do factoring by grouping.

It's not radically different from factoring by eye. Just a bit more systematic and so it can help with trinomials where the x 2 term isn't 1.
 Feb 6, 2014
 #2
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Warped Tour Girl:

Can you explain the grouping method of factoring. Can you describe a scenario when the grouping method would be preferred over other methods and show me an equation in this situation?
I need help with this, if you can please help(:



Hi Warped tour girl,
I haven't seen you for a while, welcome back.

I think this is what you want

3x+2x = (3+2)x = 5x
yes, I know that you already know that.

Let's take it a step further.
3(x-7)+2(x-7) = (3+2)(x-7)=> 5(x-7)
or
2x(x-7) + 3(x-7) = (2x+3)(x-7)

taking it back a step Lets try factorising
2x 2 - 14x + 3x - 21
I can see that it is equal to
=2x(x-7) + 3(x-7)
=(2x+3) (x-7)

The main time I would really use this is when I am factorising quadratics with a leading coefficient greater than 1
For example let's factorise this one.
2x 2-11x+5
I have to look for two numbers that multiply to 2 x 5 =10 [Since they multiply to give a pos they must have the same sign ]
and add to -11 [ add to a negative so must both be negative]
-2 and -5 don't work
-1 and -10 work beautifully -1 x -10=10 and -1+-10 = -11
now I have to split the -11x into -1x and -10x
so I have
2x 2-1x -10x+5
now I am going to factorise in pairs
x(2x-1) -5( 2x-1)
(x-5)(2x-1)
That is the main time when I would use this technique.

I actually found a video on this just yesterday that I quite like. (all except the man waving his pen around the place )
I'll go see if I can find it.
http://www.youtube.com/watch?v=ZQ-NRsWhOGI
It might be worth a watch.
 Feb 6, 2014
 #3
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Thank you soooo much, this helped tremendously. (:
 Feb 6, 2014

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