##### Factorize this problem for me

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Factor:  12x2 - 14x - 48

Oct 12th, 2015

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$\\ 12x^2-14x-48\\ \\ Take\hspace{5}2\hspace{5}as\hspace{5}common\hspace{5}factor\\ \\ =2(6x^2-7x-24)$

We have to find 2 numbers such that their product will equal

6x(-24)=-144 (which is the product of coefficient of x^2 term and the constant term)

and also their sum will equal -7(which is the coefficients of x term).

Factors of -144 are

−144,−72,−48,−36,−24,−18,−16,−12,−9,−8,−6,−4,−3,−2,−1,1,2,3,4,6,8,9,12,16,18,24,36,48,72,144

Selecting the 2 numbers from factors of 144, we get

-16 and 9

(because (-16)x9=-144 and -16+9=-7)

So now split the middle term and factorize

$\\ =2(6x^2-16x+9x-24)\\ \\ =2(2x(3x-8)+3(3x-8))\\ \\ =2(3x-8)(2x+3)\\$

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Oct 12th, 2015

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Oct 12th, 2015
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Oct 12th, 2015
Sep 26th, 2017
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