$f(x)$ must have four roots, three of which are roots of $g(x)$. Using the fact that every polynomial has a unique factorization into its roots, and since the leading coefficient of $f(x)$ and $g(x)$ are the same, we know that  $$ f(x)=g(x)(x-r) $$  where $r \in \mathbb{C}$ is the fourth root of $f(x)$. Substituting $g(x)$ and expanding, we find that  $$ \begin{aligned} f(x) \&=\left(x^{3}+a x^{2}+x+10\right)(x-r) \textbackslash\textbackslash{} \&=x^{4}+(a-r) x^{3}+(1-a r) x^{2}+(10-r) x-10 r \end{aligned} $$  Comparing coefficients with $f(x)$, we see that  $$ \begin{aligned} a-r \&=1 \textbackslash\textbackslash{} 1-a r \&=b \textbackslash\textbackslash{} 10-r \&=100 \textbackslash\textbackslash{} -10 r \&=c . \end{aligned} $$  Let\textquotesingle s solve for $a, b, c$, and $r$. Since $10-r=100, r=-90$, so $c=(-10)(-90)=900$. Since $a-r=1, a=-89$. Then, since $b=1-a r, b=-8009$. Thus, we know that $$ f(x)=x^{4}+x^{3}-8009 x^{2}+100 x+900 . $$ Taking $f(1)$, we find that $$ \begin{aligned} f(1) \&=1^{4}+1^{3}-8009(1)^{2}+100(1)+900 \textbackslash\textbackslash{} \&=1+1-8009+100+900 \textbackslash\textbackslash{} \&=(\mathbf{C})-7007 . \end{aligned} $$ 
For certain real numbers $a, b$, and $c$, the polynomial $$ g(x)=x^{3}+a x^{2}+x+10 $$ has three distinct roots, and each root of $g(x)$ is also a root of the polynomial $$ f(x)=x^{4}+x^{3}+b x^{2}+100 x+c . $$ What is $f(1)$? 
a. $$-9009$$ 
b. $$-8008$$ 
c. $$-7007$$ 
d. $$-6006$$ 
e. $$-5005$$ 