How do you determine the binomial factors of #x^3-5x^2+2x+8#?

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Answer 1 · 2016-12-28T21:31:36

#x^3-5x^2+2x+8 = (x+1)(x-4)(x-2)#

Given:

#x^3-5x^2+2x+8#

Notice that if we reverse the signs on the coefficients of the terms with odd degree then the sum of the coefficients is zero.

That is:

#-1-5-2+8 = 0#

Hence #x=-1# is a zero and #(x+1)# a factor:

#x^3-5x^2+2x+8 = (x+1)(x^2-6x+8)#

To factor the remaining quadratic note that #6=4+2# and #8=4*2#, so we find:

#x^2-6x+8 = (x-4)(x-2)#

Putting it all together:

#x^3-5x^2+2x+8 = (x+1)(x-4)(x-2)#