How do you factor #x^2+3x+4x+12#?

1 Answer
Meave60
May 11, 2015

The factors of #x^2+7x+12# are #(x+3)(x+4)#.

Factor #x^2+3x+4x+12#.

Simplify #3x+4x# to #7x#.

#x^2+7x+12#

Find two numbers that when added will make 7, and when multiplied will make 12. Numbers 3 and 4 make 7 when added and 12 when multiplied.

#(x+3)(x+4)=x^2+7x+12#

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