Question

question
factor the expression completely.
$x^{3}y^{5}-x^{4}y$
answer attempt 1 out of 2
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Explanation:

Step1: Identify greatest common factor

The GCF of $x^3y^5$ and $x^4y$ is $x^3y$.

Step2: Factor out the GCF

$x^3y^5 - x^4y = x^3y(y^4 - x)$

Step3: Factor difference of squares

$y^4 - x = (y^2)^2 - (\sqrt{x})^2$ is not applicable, but $y^4 - x$ can be written as $(y^2 - \sqrt{x})(y^2 + \sqrt{x})$, but over integers, we stop at $x^3y(y^4 - x)$. Wait, no, actually $y^4 - x$ can't be factored over integers, so the complete factorization over integers is:
$x^3y(y^4 - x) = x^3y(y^2 - \sqrt{x})(y^2 + \sqrt{x})$ but if we are factoring over polynomials with integer coefficients, we stop at $x^3y(y^4 - x)$. Wait, no, actually $y^4 - x$ is irreducible over integers. So the complete factorization is:
$x^3y(y^4 - x) = x^3y(y^2 - \sqrt{x})(y^2 + \sqrt{x})$ but if we are factoring over real numbers, we can go further, but usually, unless specified, we factor over integers. So the complete factorization over integers is $x^3y(y^4 - x)$. Wait, no, actually $y^4 - x$ can be written as $(y^2)^2 - x$, which is a difference of squares if we consider $x$ as $(\sqrt{x})^2$, but that's over real numbers. So the complete factorization over real numbers is $x^3y(y^2 - \sqrt{x})(y^2 + \sqrt{x})$, but over integers, it's $x^3y(y^4 - x)$. But let's check the original expression: $x^3y^5 - x^4y = x^3y(y^4 - x) = x^3y(y^2 - \sqrt{x})(y^2 + \sqrt{x})$. But if we are factoring over polynomials with rational coefficients, $y^4 - x$ is irreducible. So the complete factorization over integers is $x^3y(y^4 - x)$.

Wait, actually, I made a mistake. Let's start over.

Step1: Factor out GCF

The GCF of $x^3y^5$ and $x^4y$ is $x^3y$.
$x^3y^5 - x^4y = x^3y(y^4 - x)$

Step2: Check if $y^4 - x$ can be factored

Over integers, $y^4 - x$ is irreducible because it's a polynomial in two variables and can't be written as a product of non-constant polynomials with integer coefficients. So the complete factorization is $x^3y(y^4 - x)$.

Wait, but maybe we can factor $y^4 - x$ as $(y^2)^2 - x$, which is a difference of squares if we consider $x$ as $(\sqrt{x})^2$, but that's over real numbers. So if we are factoring over real numbers, we can write it as $x^3y(y^2 - \sqrt{x})(y^2 + \sqrt{x})$, but over integers, we stop at $x^3y(y^4 - x)$.

But usually, unless specified, we factor over integers. So the complete factorization is $x^3y(y^4 - x)$.

Wait, no, actually, $y^4 - x$ can be factored as $(y^2 - \sqrt{x})(y^2 + \sqrt{x})$ over real numbers, but that's not necessary unless specified. So the complete factorization over integers is $x^3y(y^4 - x)$.

Answer:

$x^3y(y^4 - x)$

Wait, actually, I think I was wrong. Let's check again. The original expression is $x^3y^5 - x^4y$. Factor out $x^3y$: $x^3y(y^4 - x)$. Then $y^4 - x$ can be written as $y^4 - x = (y^2)^2 - (\sqrt{x})^2 = (y^2 - \sqrt{x})(y^2 + \sqrt{x})$, but that's over real numbers. So if we are factoring over real numbers, that's the complete factorization, but over integers, it's $x^3y(y^4 - x)$.

But the question says "factor completely", which usually means over integers unless specified otherwise. So the answer is $x^3y(y^4 - x)$.

Wait, but maybe I made a mistake. Let's expand $x^3y(y^4 - x)$: $x^3y \cdot y^4 - x^3y \cdot x = x^3y^5 - x^4y$, which matches the original expression. So that's correct.

Alternatively, we can write $y^4 - x$ as $-(x - y^4)$, but that's not a factorization. So the complete factorization over integers is $x^3y(y^4 - x)$.

Yes, that's correct.

Final Answer:

$x^3y(y^4 - x)$