How do you factor by grouping #x^3+3x^2-118x-120#?

1 Answer
George C.
May 10, 2015

#x^3+3x^2-118x-120#
#= (x^3+2x^2-120x) + (x^2+2x-120)#
#= (x+1)(x^2+2x-120)#
#= (x+1)((x^2 + 2x + 1) - 121)#
#= (x+1)((x+1)^2 - 11^2)#
#= (x+1)(x-10)(x+12)#

I found the first grouping by noticing that #x = -1# is a solution of #x^3+3x^2-118x-120 = 0#, so #(x+1)# is a factor.

The second grouping to #(x^2 + 2x + 1) - 121# is completing the square.

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