Question

question
factor the expression completely.
(x^{4}y^{4}+x^{5}y^{2})
answer attempt 1 out of 2

Explanation:

Step1: Identify greatest common factor

The GCF of $x^4y^4$ and $xy^2$ is $xy^2$. Factor it out:
$x^4y^4 + xy^2 = xy^2(x^3y^2 + 1)$

Step2: Recognize sum of cubes

Notice $x^3y^2 + 1$ is a sum of cubes: $(xy^{2/3})^3 + 1^3$, but rewrite to use integer exponents by adjusting:
$x^3y^2 + 1 = 1 + (x y^{2/3})^3$, alternatively, factor as sum of cubes form $a^3 + b^3=(a+b)(a^2-ab+b^2)$ where $a = x y^{2/3}$, $b=1$, but to keep integer exponents, we can express it as:
$x^3y^2 + 1 = 1 + x^3y^2 = (1 + xy^{2/3})(1 - xy^{2/3} + x^2y^{4/3})$, but to make it fully factored with rational exponents or rewrite the original expression correctly. Wait, correct approach: the original expression after GCF is $xy^2(x^3y^2 + 1)$. Since $x^3y^2 = x^3 \cdot y^2$, it cannot be factored into integer polynomials further, so the complete factorization is:

Answer:

$xy^2(x^3y^2 + 1)$

Wait, correction: if we consider that maybe the original expression was $x^4y^4 + xy^2$, the GCF is $xy^2$, so factoring that out gives $xy^2(x^3y^2 + 1)$. If we allow for non-integer exponents, but typically in basic algebra, we stop at integer polynomials. So the final answer is $xy^2(x^3y^2 + 1)$.

Wait, another check: $xy^2 \cdot x^3y^2 = x^4y^4$, $xy^2 \cdot 1 = xy^2$, which adds up to the original expression. So that is the complete factorization over integer polynomials.