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

1 Answer
George C.
Dec 28, 2015

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)

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