Question 2·25 Super-Hard SAT Advanced Math Questions·Advanced Math

To evaluate $P(1)$, replace $x$ with $1$ in both factors:

A coefficient is the number multiplying a power of $x$. Only three products make $x^2$: $x^2(1)+(mx)(nx)+(1)x^2=(mn+2)x^2$. So $b=mn+2$, not just $mn$.

Let $u=m+2$ and $v=n+2$, so $uv=360$. Substitute $m=u-2$ and $n=v-2$ into $b$:

Distribute:

Replace $uv$ with the fixed product $360$: $b=366-2(u+v)$. So making $b$ greatest means making $u+v$ smallest.

For positive factors with a fixed product, compare their sum and their gap using this identity:

Here $uv$ stays $360$, so a smaller gap gives a smaller positive sum. For a fixed product, look for the closest allowed factor pair.

A square cannot be negative, so the identity gives $u+v\ge2\sqrt{uv}=2\sqrt{360}$. Type $2\sqrt{360}$ in Desmos; it shows about $37.947$. Because $u$ and $v$ are integers, their sum must be at least $38$, whose midpoint is $19$.

Two factors equal to $19$ would multiply to $361$, not $360$. Try $u=18$, one below $19$. Type $18v_1\sim360$, where $v_1$ is the other shifted factor; Desmos reports $v_1=20$ under PARAMETERS.

Type $18+v_1$; Desmos shows $38$. The pair reaches the smallest possible sum from the bound. It also gives $m=18-2=16$ and $n=20-2=18$, both positive integers.

Type $366-2(18+v_1)$; Desmos shows $290$. Since this pair has the smallest possible sum, $290$ is the greatest possible $x^2$ coefficient. Grid in 290.