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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