If you have something like $a\cdot b - a\cdot c$, you can rewrite it as $a(b - c)$. Apply this idea to the given expression.

When you rewrite the expression inside the parentheses as $(x + 5y) - (4y - x)$, be sure to distribute the negative sign across both terms in $(4y - x)$.

After distributing the minus sign, group the $x$-terms together and the $y$-terms together to simplify the expression inside the parentheses.

In a new Desmos line, type the original expression as a function of $x$ and $y$, for example: A = (x+5y)(3x-2y) - (3x-2y)(4y-x).

Click on $x$ and $y$ in the expression to add sliders (or define x = 2 and y = 1 or any other values you like). These will let you quickly test numeric values.

For each answer choice, create a new line with its expression minus A. For example, for choice A type A1 = (3x-2y)(x+y) - A, for choice B type A2 = 6x^2 + x*y - 2y^2 - A, and similarly for the others. Then move the $x$ and $y$ sliders (or change their values) and see which line always evaluates to 0 for many different $(x, y)$ pairs. The choice whose difference stays 0 is equivalent to the original expression.

The original expression is

Both terms contain the same binomial factor $(3x - 2y)$. This is like having $a\cdot b - a\cdot c$, where $a$ is $(3x - 2y)$.

Use the distributive property in reverse: $a\cdot b - a\cdot c = a(b - c)$.

Here, let $a = (3x - 2y)$, $b = (x + 5y)$, and $c = (4y - x)$.

So the expression becomes

Now simplify

First distribute the minus sign:

Combine like terms:

• Combine $x + x$ to get $2x$.
• Combine $5y - 4y$ to get $y$.

So the bracket simplifies to $2x + y$.

Substitute the simplified result back into the factored expression:

This matches answer choice D, so $(3x - 2y)(2x + y)$ is the expression equivalent to the original one.