How do you factor #2x^3 + x^2 - 4x - 3#?

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Answer 1 · 2015-05-30T17:37:14

First notice that #x=-1# is a root of #2x^3+x^2-4x-3 = 0#.

That is: if you substitute #-1# for #x# in the polynomial then the result is #0#

So #(x+1)# is a factor of #2x^3+x^2-4x-3#.

Use synthetic division by this factor...

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

Notice that #-1# is also a root of #2x^2-x-3=0#, so we have another #(x+1)# factor...

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

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