# How do you find the roots for f(x) = 17x^15 + 41x^12 + 13x^3 - 10 using the fundamental theorem of algebra?

May 14, 2016

The FTOA does not help you find the zeros - it only tells you that this polynomial of degree $15$ has exactly $15$ Complex zeros counting multiplicity.

#### Explanation:

By roots, I will assume you mean zeros, i.e. values of $x$ for which $f \left(x\right) = 0$.

The so called fundamental theorem of algebra (FTOA) is neither fundamental nor a theorem of algebra, but what it does tell you is that any non-zero polynomial in one variable with Complex coefficients has a zero in $\mathbb{C}$. That is, there is some Complex number ${x}_{1}$ such that $f \left({x}_{1}\right) = 0$.

A simple corollary of this - often stated as part of the FTOA - is that a polynomial in one variable of degree $n > 0$ with Complex coefficients has $n$ zeros counting multiplicity, all in $\mathbb{C}$.

In our example, $f \left(x\right)$ is of degree $15$, so has $15$ zeros counting multiplicity, all in $\mathbb{C}$.

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Bonus

What else can we find out about the zeros of this $f \left(x\right)$?

Note that the coefficients of $f \left(x\right)$ have only one change of sign, so only one positive Real zero.

Note $f \left(0\right) = - 10 < 0$ and $f \left(1\right) = 17 + 41 + 13 - 10 = 61 > 0$

So the positive Real zero is in $\left(0 , 1\right)$

$f \left(- x\right) = - 17 {x}^{5} + 41 {x}^{12} - 13 {x}^{3} - 10$

has two changes of sign, so $f \left(x\right)$ may have $0$ or $2$ negative Real zeros.

We find:

$f \left(- 1\right) = - 17 + 41 - 13 - 10 = 1 > 0$

So there is a Real zero in $\left(- 1 , 0\right)$ and another Real zero in $\left(- \infty , - 1\right)$.

Any other zeros will occur in Complex conjugate pairs, since all of the coefficients of $f \left(x\right)$ are Real.

Note that all of the degrees are multiples of $3$ so the zeros are all cube roots of the zeros of:

$g \left(t\right) = 17 {t}^{5} + 41 {t}^{4} + 13 t - 10$