How do you solve $(x + 4) ( x + 2) = 15$ ?

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Answer 1 · 2016-09-15T03:29:50

$x=1$ or $x=-7$

$(x+4)(x+2)=15$

$hArrx(x+2)+4(x+2)=15$ or

$x^2+2x+4x+8=15$ or

$x^2+6x-7=0$ or

$x^2+7x-x-7=0$ or

$x(x+7)-1(x+7)=0$ or

$(x-1)(x+7)=0$

i.e. either $x-1=0$ or $x+7=0$ .

Hence $x=1$ or $x=-7$

How do you solve (x + 4) ( x + 2) = 15(x + 4) ( x + 2) = 15?