Question 252·Hard·Advanced Math

Each given factor gives an input that makes the whole polynomial zero. The factor $x-1$ is zero at $x=1$, while $x+2$ is zero at the negative input $x=-2$. So $f(1)=0$ and $f(-2)=0$.

The constant term is the part without $x$. Since $r=14$, replace $r$ to get $f(x)=x^3+3px^2+qx+14$. Both unknown coefficients remain, so use the two zero conditions together in Desmos.

Type $f(x)=x^3+3px^2+qx+14$, then add $[f(1),f(-2)]\sim[0,0]$. The $\sim$ tells Desmos to fit $p$ and $q$ so both outputs are zero. Under PARAMETERS, it shows $p=-2$ and $q=-9$. The question asks for $q$, the coefficient of $x$, not $p$. Grid in -9.

The two given factors account for two of the cubic's three linear factors. Monic means the $x^3$ coefficient is $1$, so there's no extra number out front. Write the remaining factor as $x-t$:

At $x=0$, every term containing $x$ disappears, so the given $r=14$ means $f(0)=14$. Put $0$ into the product to get the third factor's value:

Set that equal to the given constant:

Type $2t\sim14$ in Desmos; $\sim$ asks it to fit the unknown $t$. Under PARAMETERS, it shows $t=7$, so the last factor is $x-7$.

Multiply the first two factors, keeping the sign on $-1$:

Combine the $x$ terms:

Multiply by the last factor:

Distribute each term:

Combine terms with the same power of $x$:

The coefficient of $x$ is $q$, so $q=-9$. Grid in -9.