Algebra Problem The expression $x^2-4x+5$ is a factor of $ax^3+bx^2+25$. Express the sum a+b as an integer. 
Please give an explanation of how the answer 
 A: If $x^2-4x+5$ is a factor of $ax^3+bx^2+25$, then $ax^3+bx^2+25=(x^2-4x+5)(\text{something})$.  Since $ax^3+\cdots$ is a polynomial of degree 3 and $x^2-\cdots$ is a polynomial of degree 2, the "something" must be a polynomial of degree 1:
$$ax^3+bx^2+0x+25=(x^2-4x+5)(\underline{\;\;\;\;\;\;}x+\underline{\;\;\;\;\;\;})$$
Try to fill in the two blanks based the terms on the left side that don't have $a$ and $b$ in them (for example, how will $+25$ end up in the product?), then finish the multiplication to find the values of $a$ and $b$.
A: $\rm\ 0 = a\ x^3 + b\ x^2 + 25 - (x^2 - 4\ x + 5)\ (a\ x + 5) = (4\ a + b - 5)\ x^2 + (20 - 5\ a)\ x\ \Rightarrow \ a,\: b = \ldots$  
Note: $\rm\ a\ x + 5\ $ comes from comparing leading and constant coefficients.
A: $ax^3+bx^2+25=(x^2-4x+5)(ax+k)$ , $k$ = constant 
$ax(x^2-4x+5) + k(x^2-4x+5) =0$ 
$ax^3 + x^2(k-4a)+x(5a-4k)+5k=0$ 
compare to original then $5k=25 \implies k=5$ 
and $(5a-4k)=0$ so $a =4$ 
and $b= k-4a = 5 -16 =-11$
A: Since $x^2-4x+5=0$ has the roots $2 \pm i$ they must also be roots of $ax^3+bx^2+25\,$ therefore:
$$
0 = a(2+i)^3+b(2+i)^2+25=8a -6a + 4b - b + 25 + i\,(12 a -a + 4b) \\
\iff
\begin{cases}
\begin{align}
2a + 3b + 25 = 0 \\
11 a + 4b = 0
\end{align}
\end{cases}
$$
