Question 114·Medium·Advanced Math

Set $x_1=1$ in Desmos, then type $2x_1^4-5x_1^2-3$. Desmos prints $-6$. Choose $1$ instead of $0$: at $0$, you'd compare only the constant terms, and all four products would tie.

Type $(2x_1^2+3)(x_1^2-1)$. Desmos prints $0$, not $-6$, so this product is not equivalent to the original.

Type $(2x_1^2-3)(x_1^2+1)$. It prints $-2$, so this product also differs from $-6$ at $x=1$.

Type $(2x_1^2-1)(x_1^2+3)$. Desmos prints $4$, which doesn't match the original's output.

Type $(2x_1^2+1)(x_1^2-3)$. It prints $-6$. The other three choices all failed at $x=1$, so this is the only choice that can be equivalent to $2x^4-5x^2-3$. Choice B.

The even powers give $x^4=(x^2)^2$. Let $u=x^2$ so the expression becomes a quadratic in $u$, with $u^2$ as its highest power: $2x^4-5x^2-3=2u^2-5u-3$.

To factor a quadratic, split its middle coefficient into two numbers that add to it and multiply to the first coefficient times the constant. Here, $-6$ and $1$ add to $-5$ and multiply to $(2)(-3)=-6$. Split $-5u$ using that pair: $2u^2-5u-3=2u^2-6u+u-3$.

Factor $2u$ from the first two terms. The last two terms already contain $u-3$, so both groups now share it: $2u^2-6u+u-3=2u(u-3)+1(u-3)$.

Factor out the shared binomial $u-3$, the two-term expression in both groups: $2u(u-3)+1(u-3)=(2u+1)(u-3)$.

Replace $u$ with all of $x^2$ in both factors: $2x^4-5x^2-3=(2x^2+1)(x^2-3)$. That's the equivalent product. Choice B.