How do you factor #x^2 + 2x + 4x + 8# by grouping?

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Answer 1 · 2016-05-15T10:41:53

#x^2+2x+4x+8=(x+4)(x+2)#

Group the terms using parentheses:

#(x^2+2x)+(4x+8)#

Identify the common factor inside each pair of parentheses and separate it out:

#x(x+2)+4(x+2)#

Combine the coefficients of the common factor:

#(x+4)(x+2)#

I would normally simply write this:

#x^2+2x+4x+8#

#=(x^2+2x)+(4x+8)#

#=x(x+2)+4(x+2)#

#=(x+4)(x+2)#