17. Given the quadratic equation in $x$, $x^{2} + b x + c = 0$.
(1) If $b$ and $c$ are two different roots of this equation, find the values of $b$ and $c$;
(2) Let $a$ and $\beta$ be the two roots of this equation, factorize
$$
\left[x^{2} + (b+1) x + c\right]^{2} + b\left[x^{2} + (b+1) x + c\right] + c \text{. }
$$
(1) From the relationship between the roots and coefficients of a quadratic equation, we have
$$
\left\{\begin{array}{l}
b+c=-b, \\
b c=c .
\end{array}\right.
$$

If $c=0$, then $b=0$, which contradicts the condition that the roots are distinct, hence $c \neq 0$.
Solving this, we get $b=1, c=-2$.
$$
\text { (2) } \begin{aligned}
\text { Original expression }= & {\left[\left(x^{2}+b x+c\right)+x\right]^{2}+b\left[\left(x^{2}+b x\right.\right.} \\
& +c)+x]+c \\
= & \left(x^{2}-b x+c\right)^{2}+(2 x+b)\left(x^{2}+b x+c\right) \\
& +\left(x^{2}+b x+c\right) \\
= & \left(x^{2}+b x+c\right)\left(x^{2}+b x+c+2 x+b+1\right) \\
= & \left(x^{2}+b x+c\right)\left[\left(x^{2}+b x+c\right)+2 x\right. \\
& -(\alpha+\beta)+1] \\
= & (x-\alpha)(x-\beta)[(x-\alpha)(x-\beta) \\
& +(x-\alpha)+(x-\beta)+1] \\
= & (x-\alpha)(x-\beta)(x-\alpha+1)(x-\beta+1) .
\end{aligned}
$$