After you know a relationship between $p$ and $q$ from the $x^3$ term, find the coefficient of $x^2$ in the product and set it equal to 24. What equation does that give you involving $pq$?

Your equations with $p+q$ and $pq$ should give two possible values for $p$ (and $q$). Use the inequality $p<q$ to decide which value of $p$ is correct and then find $q$.

Once you know $p$ and $q$, write an expression for the coefficient of $x$ in the product and plug in your $p$ and $q$. That coefficient is $r$.

From the algebra, you should have $p+q=0$ (because the $x^3$ coefficient is 0) and $25+pq=24$, which simplifies to $pq=-1$. Keep these two equations in mind as you use Desmos.

From $p+q=0$, rewrite $q=-p$ and substitute into $pq=-1$ to get $p(-p)=-1$, or $p^2=1$. In Desmos, graph $y=x^2$ and $y=1$; the $x$-coordinates where the graphs intersect are the possible values of $p$.

From the intersection points you found, use the condition $p<q$ with $q=-p$ to decide which $x$-value to take as $p$. Once you pick $p$, set $q=-p$ (the opposite sign) to get the corresponding $q$.

In Desmos, type the expression $16p+9q$, substituting the specific numbers you chose for $p$ and $q$ (for example, by defining them as constants or just typing them directly). The value that Desmos outputs for this expression is the coefficient $r$ of $x$ in the polynomial.

Write the product $(x^2+px+9)(x^2+qx+16)$ and expand it term by term:

• $x^2 \cdot x^2$ gives $x^4$.
• $x^2 \cdot qx$ and $px \cdot x^2$ give $(p+q)x^3$.
• $x^2 \cdot 16$, $px \cdot qx$, and $9 \cdot x^2$ contribute to the $x^2$ term.
• $px \cdot 16$ and $9 \cdot qx$ contribute to the $x$ term.
• $9 \cdot 16$ is the constant term.

From the $x^3$ terms we get the coefficient $(p+q)$, but the given polynomial $x^4+24x^2+rx+144$ has no $x^3$ term, meaning its $x^3$ coefficient is 0. So we must have $p+q=0$, which implies $q=-p$. This relationship will be used in the next steps.

Now find the coefficient of $x^2$ in the product. The $x^2$ terms come from:

• $x^2 \cdot 16$ giving $16x^2$
• $px \cdot qx$ giving $pqx^2$
• $9 \cdot x^2$ giving $9x^2$

So the total $x^2$ coefficient is $16 + pq + 9 = 25 + pq$.

In the given polynomial, the coefficient of $x^2$ is $24$, so we set $25 + pq = 24$, which gives $pq = -1$.

Now you have a system: $p+q=0$ and $pq=-1$.

From $p+q=0$, we already know $q=-p$. Substitute this into $pq=-1$:

• Replace $q$ with $-p$ to get $p(-p) = -1$, so $-p^2=-1$.
• This simplifies to $p^2 = 1$, so $p=1$ or $p=-1$.

If $p=1$, then $q=-1$, but that would give $p>q$, which contradicts $p<q$. Therefore, the correct pair is $p=-1$ and $q=1$, satisfying both $p+q=0$ and $p<q$.

The $x$ terms in the product come from $px \cdot 16$ and $9 \cdot qx$:

• $px \cdot 16$ gives $16p x$.
• $9 \cdot qx$ gives $9q x$.

So the coefficient of $x$ in the product (which is $r$ in the given polynomial) is $16p + 9q$.

Substitute $p=-1$ and $q=1$: $16p + 9q = 16(-1) + 9(1) = -16 + 9 = -7$.

Thus, $r=-7$, which corresponds to answer choice B.