Once you have an equation without a fraction, focus on expanding $(x^2 - 4x + 3)(x^2 - 2x + 4)$ carefully. Write out all nine products and then combine like terms.

After expanding, you will have $x^4 - 6x^3 + p x^2 + q x + 12$ on one side and a fully expanded polynomial on the other. How can you use the fact that these are equal for all $x$ to find $p$ and $q$?

Once you know the specific values of $p$ and $q$, add them to get the final answer the question is asking for.

After you have found your values for $p$ and $q$, type the expression (x^2 - 4x + 3)(x^2 - 2x + 4) into Desmos. Use Desmos’s expansion (it will show the simplified polynomial) and compare its coefficients to those in x^4 - 6x^3 + p x^2 + q x + 12 to confirm that your $p$ matches the $x^2$ coefficient and your $q$ matches the $x$ coefficient.

Still in Desmos, type the expression (x^4 - 6x^3 + p x^2 + q x + 12)/(x^2 - 4x + 3) - (x^2 - 2x + 4), using your values for $p$ and $q$. If your work is correct, the graph of this difference should be the horizontal line $y = 0$ everywhere it is defined (there will be holes where the denominator is zero). This confirms your $p$ and $q$, and therefore your value of $p + q$, are correct.

We are told that

for all $x \ne 1, 3$ (where the denominator is not zero).

Multiply both sides by $x^{2}-4x+3$ to clear the fraction:

This must hold as an identity (same polynomial on both sides for all $x$).

Now expand $(x^{2}-4x+3)(x^{2}-2x+4)$.

First distribute $x^2$:

• $x^2 \cdot x^2 = x^4$
• $x^2 \cdot (-2x) = -2x^3$
• $x^2 \cdot 4 = 4x^2$

Then distribute $-4x$:

• $-4x \cdot x^2 = -4x^3$
• $-4x \cdot (-2x) = 8x^2$
• $-4x \cdot 4 = -16x$

Then distribute $3$:

• $3 \cdot x^2 = 3x^2$
• $3 \cdot (-2x) = -6x$
• $3 \cdot 4 = 12$

Now combine like terms:

• $x^4$: coefficient $1$
• $x^3$: $-2x^3 - 4x^3 = -6x^3$
• $x^2$: $4x^2 + 8x^2 + 3x^2 = 15x^2$
• $x$: $-16x - 6x = -22x$
• constant: $12$

So the right-hand side becomes

Now we have the identity

For two polynomials to be equal for all $x$, the coefficients of each power of $x$ must match.

Compare corresponding coefficients:

• Coefficient of $x^4$: $1$ on both sides (already matches).
• Coefficient of $x^3$: $-6$ on both sides (already matches).
• Coefficient of $x^2$: $p$ on the left, $15$ on the right, so $p = 15$.
• Coefficient of $x$: $q$ on the left, $-22$ on the right, so $q = -22$.
• Constant term: $12$ on both sides (already matches).

So $p = 15$ and $q = -22$. We now use these to answer the question.

The problem asks for $p + q$.

Using $p = 15$ and $q = -22$:

So the value of $p + q$ is $-7$.