Is it possible to factor #y=4x^3-13x-6 #? If so, what are the factors?

1 Answer
mason m
Mar 14, 2016

#4x^3-13x-6=(x-2)(2x+3)(2x+1)#

Explanation:

Since there is nothing obvious to factor out at the beginning, start by plugging in small numbers for #x# to see if you can get the equation to equal #0#.

Plugging in #x=2# gives #0#, so #x-2# is a factor.

Now divide #4x^3-13x-6# by #x-2# using polynomial long division or synthetic division. If you need help with this step, say something.

#(4x^3-13x-6)/(x-2)=4x^2+8x+3#

So, from here we know that

#4x^3-13x-6=(x-2)(4x^2+8x+3)#

So now all we need to do is factor #4x^2+8x+3#. Looking for a pair of factors of #12# whose sum is #8#, we see that #2,6# works:

#4x^2+8x+3#

#=4x^2+2x+6x+3#

#=2x(2x+1)+3(2x+1)#

#=(2x+3)(2x+1)#

Thus,

#4x^3-13x-6=(x-2)(2x+3)(2x+1)#

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