Given a root, determine $a$ and $b$.

Question

Given $p(x)= x^3 + ax^2 + bx -6$ with a zero at $x=1+i$. Determine the values of $a$ and $b$.

With the question given in its current form, would it be reasonable to assume that both $a$ and $b$ are real? Since, if so, one may use the fact that $x= 1-i$ is also a zero, and therefore find a quadratic factor and use that to proceed.

Or is there another way to proceed to answer this question without having to assume real coefficients?

Answer 1

Which says that $x^3+ax^2+bx-6$ is divisible by $$(x-1+i)(x-1-i)=(x-1)^2-i^2=x^2-2x+2$$ and since $$x^3+ax^2+bx-6=x^3-2x^2+2x+(a+2)x^2+(b-2)x-6,$$ we obtain $a+2=-3$ and $b-2=6$,

which gives $$(a,b)=(-5,8).$$

Answer 2

Let $p(x) = x^3 + ax^2 + bx - 6$. We are told that $p(1 + i) = 0$. Then, $$\begin{align} (1 + i)^3 + a(1 + i)^2 + b(1 + i) - 6 &= 0\\ (-2 + 2i) + 2ai + b(1 + i) - 6 &= 0\\ (b - 8) + (2a + b + 2)i &= 0 \end{align}$$ Hence, the real and imaginary parts need to be equal to $0$. This gives us $b = 8$, and consequently, $a = -5$.

EDIT: I just realized that this assumes that $a$ and $b$ are real... So if you decide that the question requires that (reasonable IMO), then this is the solution.

EDIT 2: With celtschk's comment, I decided to take this a bit further. Let's allow the possibility for $a,b$ to be complex. Suppose that $a = x + iy$ and $b = u + iv$ where $x,y,u,v \in \mathbb{R}$. Then, $$\begin{align} ((u + iv) - 8) + (2(x + iy) + (u + iv) + 2)i &= 0\\ (u - 2y - v - 8) + (2x + u + v + 2)i &= 0 \end{align}$$ Since all the variables in the above equation are real, we can (without loss of generality) conclude that $$\begin{align} u - 2y - 2v - 8 &= 0\\ 2x + u + v + 2&= 0 \end{align}$$

So the above equations describe the complete set of solutions to $a = x + iy$ and $b = u + iv$. Notice that if $a,b \in \mathbb{R}$, then $y = v = 0$ and the solution simplifies to $b = u = 8$ and $a = x = -5$.

Answer 3

The other answers give the solution for $a,b \in \mathbb{R}$ and, since the problem was posed at a high school, I suppose that this is the case.

If we can have $a,b \in \mathbb{C}$ than the problem has infinitely many solutions. In this case also the other two solutions can be complex numbers $u$ and $v$ not related to the given $z=1+i$ and , using Vieta's formulas, we can write the system of equations: $$ \begin{cases} -a=u+v+z\\ b=uv+uz+vz\\ 6=uvz \end{cases} $$ from which, given $z$, we can find $a$ and $b$ as functions of one of the other roots.