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anonymous
 one year ago
Factor.
x2 + 7x + 12
A.
(x – 12)(x – 1)
B.
(x – 6)(x – 2)
C.
(x + 3)(x + 4)
D.
(x – 3)(x – 4)
anonymous
 one year ago
Factor. x2 + 7x + 12 A. (x – 12)(x – 1) B. (x – 6)(x – 2) C. (x + 3)(x + 4) D. (x – 3)(x – 4)

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DecentNabeel
 one year ago
Best ResponseYou've already chosen the best response.1\[\mathrm{Factor}\:x^2+7x+12:\quad \left(x+3\right)\left(x+4\right)\]

DecentNabeel
 one year ago
Best ResponseYou've already chosen the best response.1answer is option C

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0you are a life saver

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.0there are 3+ way to factor quadratic equations leading coefficient is on in this question so better to apply *headphone* method :D dw:1437342979851:dw multiply AC then find two number if you multiply them you should get product of AC and if you addo r subtract them you should get middle term

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.0MR. Decent nabeel why so hurry eh ?

DecentNabeel
 one year ago
Best ResponseYou've already chosen the best response.1ok.. first break the expression into groups.. \[=\left(4x+12\right)+\left(x^2+3x\right)\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i appreciate the time your taking @Nnesha

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0and you to @DecentNabeel

DecentNabeel
 one year ago
Best ResponseYou've already chosen the best response.1then \[\mathrm{Factor\:out\:}4\mathrm{\:from\:}4x+12\mathrm{:\quad }4\left(x+3\right)\] \[\mathrm{Factor\:out\:}x\mathrm{\:from\:}x^2+3x\mathrm{:\quad }x\left(x+3\right)\] \[=x\left(x+3\right)+4\left(x+3\right)\] factor out x+3 \[=\left(x+4\right)\left(x+3\right)\]

DecentNabeel
 one year ago
Best ResponseYou've already chosen the best response.1@Nnesha are you ok

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oooo ok thanks a who'll lot.

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.0yes i'm thanks for asking.....

DecentNabeel
 one year ago
Best ResponseYou've already chosen the best response.1wellcoem :) @Nnesha

DecentNabeel
 one year ago
Best ResponseYou've already chosen the best response.1welcome @speartonion
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