# Thread: Polynomial with real coefficients

1. ## Polynomial with real coefficients

I have a problem that I am not understanding for homework:

Find a polynomial with real coefficients, degree of 3, zeros of 2,1,-i, and satisfies the condition that f(0)=4
I assumed that thus the constant would be '4'... (ax^3 + ax^2 +ax +4)
And I assumed that the zeros would meand: (x-2)(x-1) and (x^2+1)

The problem that I am facing is that the above would result in a degree of 4, not 3. Is there some small bit I am missing, or am I misinterperting (sp?) the question.

2. Hi, airyie!

Originally Posted by airyie
The problem that I am facing is that the above would result in a degree of 4, not 3. Is there some small bit I am missing, or am I misinterperting (sp?) the question.
This problem has no solution. If $\displaystyle a,\;b\in\mathbb{R}\text{ and }a+b\,\text{i}$ is a root of a polynomial with real coefficients in one variable, then $\displaystyle a-b\,\text{i}$ is also a root of that polynomial. This implies that, for your polynomial to have a root of $\displaystyle -\text{i}$, it must also have a root of $\displaystyle +\text{i}$. And, since the fundamental theorem of algebra states that a polynomial with complex coefficients of degree $\displaystyle n$ can have no more than $\displaystyle n$ complex roots, it follows that the polynomial sought in the problem must be of degree 4 or higher.

Now, if you want to find a fourth-degree polynomial with the given characteristics, you have the right idea:

$\displaystyle f(x) = a\left(x-2\right)\left(x-1\right)\left(x^2+1\right)$

$\displaystyle f(0) = 2a = 4 \Rightarrow a = 2$

$\displaystyle \Rightarrow f(x) = 2\left(x-2\right)\left(x-1\right)\left(x^2+1\right) = 2x^4-6x^3+6x^2-6x+4$