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#### agentmulder

##### Active member

- Feb 9, 2012

- 33

I agree with everything MarkFL has posted so i'll just give a quick 'scheme' leaving out the details.Moderator edit: This commentary topic pertains to the following tutorial:

http://mathhelpboards.com/math-notes-49/factoring-quadratics-3396.html

Another approach is to factor by grouping, consider,

$6x^2 - 17x - 45 $

$(6)(-45) = -270$

Now find 2 integers that multiply to -270 and sum to -17,

-27 with 10 work

rewrite $-17x$ as $-27x + 10x $

$6x^2 - 27x + 10x - 45 $

use grouping, paying mind to the signs,

$ (6x^2 - 27x) + (10x -45)$

factor,

$3x(2x - 9) + 5(2x - 9)$

$(2x - 9)(3x + 5)$

You can multiply this out using FOIL to confirm the factoring is correct.

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