anonymous
  • anonymous
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
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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DecentNabeel
  • DecentNabeel
\[\mathrm{Factor}\:x^2+7x+12:\quad \left(x+3\right)\left(x+4\right)\]
DecentNabeel
  • DecentNabeel
answer is option C
anonymous
  • anonymous
you are a life saver

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More answers

Nnesha
  • Nnesha
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
Nnesha
  • Nnesha
MR. Decent nabeel why so hurry eh ?
anonymous
  • anonymous
lol
DecentNabeel
  • DecentNabeel
ok.. first break the expression into groups.. \[=\left(4x+12\right)+\left(x^2+3x\right)\]
Nnesha
  • Nnesha
enabling
anonymous
  • anonymous
i appreciate the time your taking @Nnesha
anonymous
  • anonymous
and you to @DecentNabeel
anonymous
  • anonymous
(:
DecentNabeel
  • DecentNabeel
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 \[=\left(x+4\right)\left(x+3\right)\]
DecentNabeel
  • DecentNabeel
@Nnesha are you ok
anonymous
  • anonymous
oooo ok thanks a who'll lot.
Nnesha
  • Nnesha
yes i'm thanks for asking.....
DecentNabeel
  • DecentNabeel
wellcoem :) @Nnesha
DecentNabeel
  • DecentNabeel
welcome @speartonion
anonymous
  • anonymous
(:

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