# Polynomial with real coefficients

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• May 19th 2008, 05:01 PM
airyie
Polynomial with real coefficients
I have a problem that I am not understanding for homework:

Quote:

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.
• May 19th 2008, 05:27 PM
Reckoner
Hi, airyie!

Quote:

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 $a,\;b\in\mathbb{R}\text{ and }a+b\,\text{i}$ is a root of a polynomial with real coefficients in one variable, then $a-b\,\text{i}$ is also a root of that polynomial. This implies that, for your polynomial to have a root of $-\text{i}$, it must also have a root of $+\text{i}$. And, since the fundamental theorem of algebra states that a polynomial with complex coefficients of degree $n$ can have no more than $n$ 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:

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

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

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