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

1 Answer
Dec 28, 2015

Answer:

#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)#