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

Mathematics
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\[\mathrm{Factor}\:x^2+7x+12:\quad \left(x+3\right)\left(x+4\right)\]
answer is option C
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there 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
MR. Decent nabeel why so hurry eh ?
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ok.. first break the expression into groups.. \[=\left(4x+12\right)+\left(x^2+3x\right)\]
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i appreciate the time your taking @Nnesha
and you to @DecentNabeel
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then \[\mathrm{Factor\:out\:}4\mathrm{\:from\:}4x+12\mathrm{:\quad }4\left(x+3\right)\] \[\mathrm{Factor\:out\:}x\mathrm{\:from\:}x^2+3x\mathrm{:\quad }x\left(x+3\right)\] \[=x\left(x+3\right)+4\left(x+3\right)\] factor out x+3 \[=\left(x+4\right)\left(x+3\right)\]
@Nnesha are you ok
oooo ok thanks a who'll lot.
yes i'm thanks for asking.....
wellcoem :) @Nnesha
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