Question 33·Medium·Advanced Math

Choose $t=1$ because the terms containing $t$ won't disappear. In Desmos, let $x_1$ stand for that input. Type $x_1=1$, then type the two given formulas with $x_1$ in place of $t$. Desmos gives $a=-1$ and $b=4$, so both given definitions are accounted for.

Add $a^2-ab+2b$. Desmos shows $13$. The sign matters: with $a=-1$ and $b=4$, $ab$ is negative, so subtracting $ab$ adds to the total.

Type the three expressions shown in the list, in that order. Desmos returns $[15,3,27]$. Each differs from the original's $13$ at $t=1$, so none can be equivalent. An equivalent expression must match at every input, including this one.

Type the remaining expression with $x_1$ in place of $t$. Desmos shows $13$, matching the original. A match at one input isn't a proof on its own, but every other choice has failed. So $a^2-ab+2b$ is equivalent to $6t^2-27t+34$. Choice C.

A square means multiplying two copies. Use the distributive property: multiply each term in the first copy by each term in the second, then combine the middle terms:

Multiply every term in $2t-3$ by every term in $5-t$. In particular, $(-3)(-t)=+3t$:

The $2$ multiplies both terms of $b=5-t$. Distribute it:

Put the three expanded pieces into $a^2-ab+2b$. Then flip every sign in $ab$, because the expression subtracts that whole product:

Like terms have the same variable part. Group the $t^2$ terms, the $t$ terms, and the constants, then add each group: $\begin{aligned}a^2-ab+2b&=(4t^2+2t^2)+(-12t-13t-2t)+(9+15+10)\\&=6t^2-27t+34.\end{aligned}$ This holds for every $t$, confirming the equivalent expression. Choice C.