# How do you factor completely x^3y^2+x?

Dec 28, 2015

${x}^{3} {y}^{2} + x = x \left({x}^{2} {y}^{2} + 1\right) = x \left(x y - i\right) \left(x y + i\right)$

#### Explanation:

First notice that both of the terms are divisible by $x$, so separate that out as a factor first:

${x}^{3} {y}^{2} + x = x \left({x}^{2} {y}^{2} + 1\right)$

Next note that if $x$ and $y$ are Real numbers then ${x}^{2} \ge 0$, ${y}^{2} \ge 0$ and so ${x}^{2} {y}^{2} + 1 \ge 1 > 0$. So there are no linear factors with Real coefficients.

The difference of squares identity can be written:

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

If we permit Complex coefficients, then we can use ${i}^{2} = - 1$ and the difference of squares identity to find:

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

Putting this all together we have:

${x}^{3} {y}^{2} + x = x \left({x}^{2} {y}^{2} + 1\right) = x \left(x y - i\right) \left(x y + i\right)$