Try letting $y = x^2$. What does $2x^4 - 5x^2 - 3$ become in terms of $y$?

Once you have $2y^2 - 5y - 3$, look for two numbers that multiply to $-6$ and add to $-5$. Use them to factor by grouping.

After factoring in terms of $y$, replace $y$ with $x^2$ again. Then see which answer choice matches your factored expression.

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

Type the four options as separate functions:

• g(x) = (2x^2 + 3)(x^2 - 1)
• h(x) = (2x^2 + 1)(x^2 - 3)
• j(x) = (2x^2 - 3)(x^2 + 1)
• k(x) = (2x^2 - 1)(x^2 + 3)

Look at how each of the graphs $g(x)$, $h(x)$, $j(x)$, and $k(x)$ compares to $f(x)$. The function whose graph lies exactly on top of $f(x)$ for all $x$ (they are indistinguishable) is the expression that is equivalent to $2x^4 - 5x^2 - 3$.

Notice that the expression

has powers $x^4$, $x^2$, and a constant. This means you can treat it like a quadratic in $x^2$ by letting $y = x^2$, so the expression becomes

Now factor $2y^2 - 5y - 3$.

We need two numbers that multiply to $2 \cdot (-3) = -6$ and add to $-5$. Those numbers are $-6$ and $1$.

Rewrite the middle term using $-6y$ and $+y$:

Group and factor:

Replace $y$ with $x^2$ to go back to the original variable:

So the expression equivalent to $2x^4 - 5x^2 - 3$ is $(2x^2 + 1)(x^2 - 3)$, which corresponds to choice B.