Is it possible to factor #y=x^3 + 4x^2 - x #? If so, what are the factors?

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Answer 1 · 2015-12-20T11:58:30

Yes, with irrational coefficients:

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

Separate out the common factor #x# of the terms, complete the square then use the difference of squares identity to finish.

The difference of squares identity can be written:

#a^2-b^2=(a-b)(a+b)#

We use this below with #a=(x+2)# and #b=sqrt(5)#

So we find:

#x^3+4x^2-x#

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

#=x(x^2+4x+4-5)#

#=x((x+2)^2-(sqrt(5))^2)#

#=x((x+2)-sqrt(5))((x+2)+sqrt(5))#

#=x(x+2-sqrt(5))(x+2+sqrt(5))#