# How do you write a polynomial with function of minimum degree in standard form with real coefficients whose zeros include -3,4, and 2-i?

Dec 19, 2015

$P \left(X\right) = a q \left(X + 3\right) \left(X - 4\right) \left(X - 2 + i\right) \left(X - 2 - i\right)$ with $a q \in \mathbb{R}$.

#### Explanation:

Let $P$ be the polynomial you're talking about. I assume $P \ne 0$ or it would be trivial.

P has real coefficients, so $P \left(\alpha\right) = 0 \implies P \left(\overline{\alpha}\right) = 0$. It means that there is another root for P, $\overline{2 - i} = 2 + i$, hence this form for $P$ :

$P \left(X\right) = a {\left(X + 3\right)}^{{a}_{1}} \cdot {\left(X - 4\right)}^{{a}_{2}} \cdot {\left(X - 2 + i\right)}^{{a}_{3}} \cdot {\left(X - 2 - i\right)}^{{a}_{4}} \cdot Q \left(X\right)$ with ${a}_{j} \in \mathbb{N}$, $Q \in \mathbb{R} \left[X\right]$ and $a \in \mathbb{R}$ because we want $P$ to have real coefficients.

We want the degree of $P$ to be as small as possible. If $R \left(X\right) = a {\left(X + 3\right)}^{{a}_{1}} {\left(X - 4\right)}^{{a}_{2}} {\left(X - 2 + i\right)}^{{a}_{3}} {\left(X - 2 - i\right)}^{{a}_{4}}$ then $\mathrm{de} g \left(P\right) = \mathrm{de} g \left(R\right) + \mathrm{de} g \left(Q\right) = \sum \left({a}_{j} + 1\right) + \mathrm{de} g \left(Q\right)$. $Q \ne 0$ so $\mathrm{de} g \left(Q\right) \ge 0$. If we want $P$ to have the smallest degree possible, then $\mathrm{de} g \left(Q\right) = 0$ ($Q$ is just a real number $q$), hence $\mathrm{de} g \left(P\right) = \mathrm{de} g \left(R\right)$ and here we can even say that $P = R$. $\mathrm{de} g \left(P\right)$ will be as small as possible if each ${a}_{j} = 0$. So $\mathrm{de} g \left(P\right) = 4$.

So for now, $P \left(X\right) = a \left(X + 3\right) \left(X - 4\right) \left(X - 2 + i\right) \left(X - 2 - i\right) q$. Let's develop that.

$P \left(X\right) = a q \left({X}^{2} - X - 12\right) \left({X}^{2} - 4 X + 5\right) \in \mathbb{R} \left[X\right]$. So this expression is the best $P$ we can find with those conditions!