How do you factor $(x+2)^2-7(x+2)+12$ ?

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Answer 1 · 2015-11-10T20:00:53

$(x-1)(x-2)$

Expand all the terms:

$(x^2+4x+4) - (7x-14) + 12 = x^2 - 3x + 2$

Solve the quadratic equation: when dealing with a quadratic $ax^2+bx+c$ , we know that the two solutions are given by the formula

$x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

Plugging the values,

$x_{1,2} = \frac{3\pm\sqrt{9-8}}{2} = \frac{3\pm1}{2}$

Which means that $x_1=1$ and $x_2=2$ .

When you find the solutions of a quadratic equation, if $a=1$ , you can write

$x^2+bx+c=(x-x_1)(x-x_2)$ , so in your case

$x^2 - 3x + 2 = (x-1)(x-2)$

How do you factor (x+2)^2-7(x+2)+12(x+2)^2-7(x+2)+12?