Find coefficients of a cubic function with imaginary root I have a cubic function $ax^3+bx^2+cx+d$ with roots at $x=2$ and $x=1+4i$. There is also a condition that $f(1)=-32$. I have to find $a, b, c, d$ with just these three pieces of information.
Because when you take the root of something negative, you get both +/- of that answer, can I assume that another solution is $x=1-4i$?
If this is the case, how would I use these solutions to find the coefficients of the function? If $x=1-4i$ is not a root, how would I go about solving for four different values with only three constraints to work with?
Thanks!
 A: As hinted in a comment, we assume here that the coefficients $a,b,c,d$ are real.
If $x=1+4i $ is a root, then also its conjugate $x=1-4i$. Now here is where I differ.
It is better to write $x-1=4i$, then square which gives $x^2-2x+1=-16. $ Do you recognize that this move captures the conjugate root as well? So we have $x^2-2x+17$ as one part of the polynomial. The other root $x=2$ comes from the factor $x-2$ and so we arrive at $f(x)=a(x-2)(x^2-2x+17). $ Now substituting $(1,-32)$ for $x$ and $y$ gives you access to find $a$. At this point you can work out the brackets if needed, but that's basic algebra. 
A: Assuming the coefficients are real (otherwise the problem would be underdetermined), you know that also $1-4i$ is a root and, from Viète's formulas, that
$$
\begin{cases}
-\dfrac{b}{a}=2+1+4i+1-4i=4 \\[6px]
\dfrac{c}{a}=2(1+4i)+2(1-4i)+(1+4i)(1-4i)=21 \\[6px]
-\dfrac{d}{a}=2(1+4i)(1-4i)=34
\end{cases}
$$
Hence $b=-4a$, $c=21a$ and $d=-34a$. The last relation is $f(1)=-32$, so
$$
a+b+c+d=-32
$$
