Distribute the minus sign across $x^2-5x+4$, then combine like terms to rewrite $p(x)-q(x)$ in the form $Ax^2+Bx+C$ involving $a$.

Once you have $p(x)-q(x)$ simplified, set it equal to $4x^2+8x-2$. Compare the coefficients of $x^2$, $x$, and the constant term on both sides. Which equation from these coefficients involves $a$?

Only one of the coefficients (of $x^2$, $x$, or the constant) contains $a$. Use that coefficient equation to solve for $a$ directly.

Type p(x) = a x^2 + 3x + 2 in Desmos; accept the slider for a. Then type q(x) = x^2 - 5x + 4.

Add a new line and type p(1) - q(1); Desmos will display this as an expression in a. On another line, type 4(1)^2 + 8(1) - 2 to see the corresponding value from $4x^2+8x-2$ at $x=1$.

Now type the equation p(1) - q(1) = 4(1)^2 + 8(1) - 2. Desmos will solve this equation and show the value of a that makes it true; that value is the correct choice.

We are given

Compute $p(x)-q(x)$ by treating $q(x)$ as a grouped expression:

The parentheses around $q(x)$ are important so the minus sign is distributed to every term of $q(x)$.

Distribute the negative sign in $-(x^{2}-5x+4)$:

Now combine like terms:

• For $x^2$: $ax^2 - x^2 = (a-1)x^2$
• For $x$: $3x + 5x = 8x$
• Constants: $2 - 4 = -2$

So

We are told that

From Step 2 we also have

Because these expressions are equal for all $x$, their corresponding coefficients must match:

• Coefficient of $x^2$: $a-1$ must equal $4$.
• Coefficient of $x$: $8$ must equal $8$ (already true).
• Constant term: $-2$ must equal $-2$ (already true).

Use the $x^2$-coefficient equation:

Add $1$ to both sides:

So the value of $a$ is $5$, which corresponds to choice C.