After expanding, group the $x^{3}$, $x^{2}$, $x$, and constant terms separately so you can clearly see the coefficient of each power of $x$.

Set the coefficient of $x^{2}$ in your expanded expression equal to $-7$, the coefficient of $x$ equal to $5$, and the constant term equal to $12$, then solve the resulting equations for $p$, $q$, and $r$.

Once you find $p$, $q$, and $r$, the problem asks for $p+q+r$, not for the individual values themselves.

Type f(x) = (x^2 + p x + q)(2x - 3) + r into Desmos. When prompted, add sliders for $p$, $q$, and $r$. Then type g(x) = 2x^3 - 7x^2 + 5x + 12 as a second function.

Adjust the sliders for $p$, $q$, and $r$ until the graph of $f(x)$ lies exactly on top of the graph of $g(x)$ for all visible $x$-values. When the graphs coincide perfectly, note the values of $p$, $q$, and $r$ shown by the sliders.

Use the $p$, $q$, and $r$ values from the sliders and calculate $p+q+r$ (you can type this as an expression in Desmos or compute it by hand) to obtain the final numerical result.

First ignore $r$ and expand $(x^{2}+px+q)(2x-3)$.

Now distribute in each term:

Group the expanded terms by powers of $x$ and then add $r$:

• For $x^{3}$: $2x^{3}$
• For $x^{2}$: $-3x^{2} + 2px^{2} = (-3 + 2p)x^{2}$
• For $x$: $-3px + 2qx = (-3p + 2q)x$
• Constant: $-3q$ (and then $+r$)

So the whole expression becomes

We are told this expression is equivalent to

for all real $x$. That means the coefficients of the same powers of $x$ must be equal.

Match term by term:

• $x^{3}$: $2$ (already matches, gives no new info)
• $x^{2}$: $-3 + 2p = -7$
• $x^{1}$: $-3p + 2q = 5$
• Constant: $-3q + r = 12$

Now solve these equations one by one.

From the $x^{2}$ coefficients:

Add $3$ to both sides:

Use $p = -2$ in the $x$-coefficient equation:

So

Now use $q = -\tfrac{1}{2}$ in the constant-term equation:

So

Now add the three values:

• $p = -2$
• $q = -\tfrac{1}{2}$
• $r = \tfrac{21}{2}$

Compute the sum:

Convert $-2$ to halves: $-2 = -\tfrac{4}{2}$, so

So $p + q + r = 8$.