How do you factor # x^3+8x^2+19x+12#?

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Answer 1 · 2016-04-27T21:49:03

#x^3+8x^2+19x+12 = (x+1)(x+3)(x+4)#

#f(x) = x^3+8x^2+19x+12#

Note that the coefficients of #f(-x) = -x^3+8x^2-19x+12# add up to #0#.

So #f(-1) = 0# and #(x+1)# is a factor:

#x^3+8x^2+19x+12 = (x+1)(x^2+7x+12)#

Then notice that #3+4=7# and #3xx4#=12#

Hence:

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

Putting it all together:

#x^3+8x^2+19x+12 = (x+1)(x+3)(x+4)#