How do you factor #x^3 - 7x - 6#?

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Answer 1 · 2017-02-25T22:10:30

#x^3-7x-6 = (x-1)(x-3)(x+2)#

Given:

#x^3-7x-6#

Notice that if you reverse the signs of the coefficients on the terms of odd degree, then their sum is #0#. That is:

#-1+7-6 = 0#

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

#x^3-7x-6 = (x+1)(x^2-x-6)#

To factor the remaining quadratic, find a pair of factors of #6# which differ by #1#. The pair #3, 2# works and hence we find:

#x^2-x-6 = (x-3)(x+2)#

Putting it all together:

#x^3-7x-6 = (x-1)(x-3)(x+2)#