Remainder‑Theorem / Synthetic‑Division Technique

Solution (one‑minute route)

The quadratic divisor factors as $(x-1)(x-2)$. For any polynomial $P$,

so the remainder $ax+b$ must match $P(x)$ at the two roots $x=1$ and $x=2$:

$$ \begin{aligned} P(1) &= a(1)+b = a+b,\ P(2) &= a(2)+b = 2a+b. \end{aligned} $$

Compute the two evaluations (synthetic division isn’t even needed):

$$ \begin{aligned} P(1) &= 1^{4}+4\cdot1^{3}-5\cdot1+7 = 1+4-5+7 = 7,\ P(2) &= 2^{4}+4\cdot2^{3}-5\cdot2+7 = 16+32-10+7 = 45. \end{aligned} $$

Now solve the linear system

$$ \begin{cases} a+b=7,\ 2a+b=45. \end{cases} \quad\Longrightarrow\quad a=38,; b=-31. $$

Finally, the problem asks for $a+b = 38 + (-31) =$ $\boxed{7}$