After substituting $y = x^2$, you get a quadratic in $y$. How do you factor $2y^2 - 5y - 3$? Think about two numbers that multiply to $-6$ and add to $-5$.

Once you factor the quadratic in $y$, replace $y$ with $x^2$ again. Then see which answer choice has that same factored form.

In Desmos, type f(x) = 2x^4 - 5x^2 - 3 to define the original function.

For each option, type it as a new function, for example:

• g(x) = (2x^2 + 1)(x^2 - 3)
• h(x) = (2x^2 - 1)(x^2 + 3)
• p(x) = (2x^2 - 3)(x^2 + 1)
• q(x) = (2x^2 + 3)(x^2 - 1)

Either:

• Look at the graph and see which function’s curve lies exactly on top of $f(x)$ for all visible $x$, or
• Open a table for $f(x)$ and each choice and compare their $y$-values at several $x$-values (like $x = -2, -1, 0, 1, 2$).

The choice whose function always matches $f(x)$ is the equivalent expression.

Notice the powers: the expression is $2x^4 - 5x^2 - 3$. The exponents are $4$ and $2$, so you can view this as a quadratic in $x^2$.

Let $y = x^2$. Then $x^4 = (x^2)^2 = y^2$, so the expression becomes:

Now you just need to factor this quadratic in $y$.

Factor $2y^2 - 5y - 3$.

Look for two numbers that:

• multiply to $2 \cdot (-3) = -6$
• add to $-5$

Those numbers are $-6$ and $1$.

Use them to factor by grouping:

Now replace $y$ with $x^2$ to express the factors in terms of $x$. You'll compare this factored form to the answer choices in the next step.

Compare the factored expression you found, $(2x^2 + 1)(x^2 - 3)$, to the answer choices.

It exactly matches choice A, so the expression equivalent to $2x^4 - 5x^2 - 3$ is $(2x^2 + 1)(x^2 - 3)$.