# How do you find the number of roots for f(x) = 4x^5 – 2x – 15 using the fundamental theorem of algebra?

Mar 23, 2016

$f \left(x\right)$ has exactly $5$, possibly Complex, zeros counting multiplicity.

#### Explanation:

The Fundamental Theorem of Algebra tells us that any non-constant polynomial equation with Complex (possibly Real) coefficients has at least one Complex (possibly Real) zero.

A simple corollary of this is that a polynomial of degree $n > 0$ with Complex coefficients has exactly $n$ zeros in $\mathbb{C}$ counting multiplicity.

To show the corollary: Suppose $p \left(x\right)$ is a polynomial of degree $n > 0$. Then $p \left(x\right)$ has some zero ${r}_{1} \in \mathbb{C}$. Then $\left(x - {r}_{1}\right)$ is a factor of $p \left(x\right)$. So $p \left(x\right) = \left(x - {r}_{1}\right) q \left(x\right)$ for some polynomial $q \left(x\right)$ of degree $n - 1$. If $n - 1 > 0$ then $q \left(x\right)$ has a zero ${r}_{2} \in \mathbb{C}$ and $q \left(x\right)$ is divisible by $\left(x - {r}_{2}\right)$. Repeat until we have all $n$ zeros ${r}_{1} , {r}_{2} , \ldots , {r}_{n} \in \mathbb{C}$.

So in the case of $f \left(x\right) = 4 {x}^{5} - 2 x - 15$ we can deduce that $f \left(x\right)$ has exactly $5$ (possibly Complex) zeros counting multiplicity.

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Bonus

What else can we find out about these zeros?

First note that the sequence of coefficients of $f \left(x\right)$ has one change of sign, so $f \left(x\right)$ has at most one positive Real zero.

In fact, we can see that $f \left(1\right) = 4 - 2 - 15 = - 13 < 0$ and $f \left(2\right) = 128 - 4 - 15 = 109 > 0$, so $f \left(x\right)$ has a zero in $\left(1 , 2\right)$.

If we reverse the signs of the coefficients of the terms of odd degree, then all are negative. So $f \left(x\right)$ has no negative Real roots.

Since the coefficients of $f \left(x\right)$ are all Real, any non-Real Complex zeros must occur in conjugate pairs.

So we must have two pairs of Complex conjugate zeros and one positive Real zero.