Question

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

Answer 1

`x^3y^2+x = x(x^2y^2+1) = x(xy-i)(xy+i)`

Explanation:

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

`x^3y^2+x = x(x^2y^2+1)`

Next note that if `x` and `y` are Real numbers then `x^2 >= 0`, `y ^2 >= 0` and so `x^2y^2 + 1 >= 1 > 0`. So there are no linear factors with Real coefficients.

The difference of squares identity can be written:

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

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

`x^2y^2+1 = (xy)^2 - i^2 = (xy-i)(xy+i)`

Putting this all together we have:

`x^3y^2+x = x(x^2y^2+1) = x(xy-i)(xy+i)`