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

Dec 20, 2015

Yes, with irrational coefficients:

${x}^{3} + 4 {x}^{2} - x = x \left(x + 2 - \sqrt{5}\right) \left(x + 2 + \sqrt{5}\right)$

#### Explanation:

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:

${a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)$

We use this below with $a = \left(x + 2\right)$ and $b = \sqrt{5}$

So we find:

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

$= x \left({x}^{2} + 4 x - 1\right)$

$= x \left({x}^{2} + 4 x + 4 - 5\right)$

$= x \left({\left(x + 2\right)}^{2} - {\left(\sqrt{5}\right)}^{2}\right)$

$= x \left(\left(x + 2\right) - \sqrt{5}\right) \left(\left(x + 2\right) + \sqrt{5}\right)$

$= x \left(x + 2 - \sqrt{5}\right) \left(x + 2 + \sqrt{5}\right)$