maths factorising

mufc802

mufc802

mufc802

Soldato
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1 Jul 2009
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I'm a little confused can someone explain why this works..

ep-fp+eq-fq : Finding the common factors..

=p(e-f)+q(e-f) : and putting the common factors outside each bracket

=(p+q)(e-f) So why do both outside the bracket end up in the same bracket?

I'm just a little confused.:p
 
(e-f) is a common factor. When you factorise you're left with (p+q).

I think you're getting confused with the algebra. It's better to break things down, there is no shame in doing it the simple way.



Let's call (e-f) = X.

If this is the case then from your equation p(e-f)+q(e-f), it can be written as pX+qX.

You know X is repeated as the common factor, it's easy to take this outside and say X(p+q).

By this point alarm bells should be ringing. We originally said (e-f) = X, so then it's also true to write (p+q)(e-f).
 
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2x4 - 3x4 + 2x5 - 3x5 = 4(2-3) + 5(2-3) = (4+5)(2-3) = -9

Seems to work :D

I guess the bracketerising is down to the commonness. I can't remember that far back.
 
I think I get that now, but can't seem to get the common factors in:
6xy + 9x + 4y + 6..

6xy & 4y = y(6x+4) (edit: ah i see the common factor is actually 2y) 2y(3x+2)
9x & 6 = 3(3x+2) and their both different in the brackets?

= (2y+3)(3x+2):)
 
Last edited:
Yateszy said:
=(p+q)(e-f) So why do both outside the bracket end up in the same bracket?
Click to expand...

You're just following the same method as the first factorisation that you do "get". Like Ricochet says, just ignore the complicated looking "(e-f)" and pretend it's something simpler, if that makes it easier to see.:)
 
Yateszy said:
I think I get that now, but can't seem to get the common factors in:
6xy + 9x + 4y + 6..

6xy & 4y = y(6x+4) (edit: ah i see the common factor is actually 2y) 2y(3x+2)
9x & 6 = 3(3x+2) and their both different in the brackets?

= (2y+3)(3x+2):)
Click to expand...

Well done! You're getting the hang of it now.
 
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