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)#