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:

$x^3+4x^2-x$

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

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

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

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

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

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