# Thread: Polynomial function with imaginary root

1. ## Polynomial function with imaginary root

The polynomial 3x^3-13x^2+ax+b=0, a and b real numbers, has 2-i as one of its roots. Where does the graph of y=3x^3-13x^2+ax+b cross the x-axis?

The solution is 1/3. Through what method(s) could i find that answer?

2. Originally Posted by ceinstein1
has 2-i as one of its roots.
One of the other roots is its complex conjugate 2+i, then use these with the fact that if P(a) = 0 then (x-a) is a factor.

Also if you want to graph this you with have to use a complex plane with an imaginary and real axis.

3. If $\displaystyle 2-i$ is a root of $\displaystyle f(x) = 3x^3-13x^2+ax+b$, then so is $\displaystyle 2+i$.
$\displaystyle (x-[2-i])(x-[2+i]) = x^2-4x+5$, so we can write
$\displaystyle 3x^3-13x^2+ax+b = (x^2-4x+5)(cx+d)$ for some $\displaystyle c, d \in\mathbb{R}$.
By expanding the RHS, or otherwise, we see from the coefficient of $\displaystyle x^3$ that $\displaystyle c = 3$,
and from that of $\displaystyle x^2$ we have $\displaystyle d-4c = -13$, and as $\displaystyle c = 3$, we have $\displaystyle d = -1$.
Thus it crosses the x-axis at $\displaystyle x = \frac{1}{3}.$