Question 180·Hard·Advanced Math

No Desmos needed. An excluded denominator zero reveals the requested sum directly. Multiply both sides by $x^2-4x+3$ at the allowed inputs:

This is a polynomial identity: two polynomials that agree at infinitely many inputs agree everywhere. So this new equation also holds at $x=1$, even though the original fraction is undefined there.

The excluded input $1$ makes the denominator factor $x^2-4x+3$ equal $0$, so the whole right side of the identity is $0$. On the left, $p(1)^2+q(1)$ gives the sum you need. Substitute into the identity, not the original fraction:

Each power of $1$ is $1$, so the unknown terms are now $p$ and $q$:

Combine the known numbers $1-6+12$:

Subtract $7$ from both sides to isolate $p+q$:

So the requested sum is $-7$. Grid in -7.

For a more general identity with no useful denominator zero, you can fit the unknown constants. Type $x_1=[2,4,5]$ into Desmos. It shows the list $[2,4,5]$; none is excluded, so the original fraction is defined at each input.

Type the given equivalence with $x_1$ in place of $x$ and $\sim$ in place of $=$. The $\sim$ tells Desmos to fit the unknown constants using your list. Under PARAMETERS, it shows $p=15$ and $q=-22$.

The question asks for their sum, not either constant alone. Type $p+q$ on a new line; Desmos prints $-7$. So $p+q=-7$. Grid in -7.