# How do you find the roots for f(x)=16x^4+8x^3+4x^2-2x-15 using the fundamental theorem of algebra?

Aug 20, 2016

The FTOA tells us that there are $4$ zeros, but not how to find them.

#### Explanation:

Fundamental Theorem of Algebra

The fundamental theorem of algebra (FTOA) tells us that any non-constant polynomial with Complex (possibly Real) coefficients will have a zero in the Complex numbers.

A straightforward corollary of that, often stated as part of the FTOA is that a polynomial of degree $n > 0$ has exactly $n$ Complex (possibly Real) zeros counting multiplicity.

Given:

$f \left(x\right) = 16 {x}^{4} + 8 {x}^{3} + 4 {x}^{2} - 2 x - 15$

Note that $f \left(x\right)$ is of degree $4$. So by the FTOA we deduce that it has $4$ Complex (possibly Real) zeros counting multiplicity.

That is all the FTOA tells us. It does not tell us how to find them.

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Rational Roots Theorem

By the rational roots theorem, any rational zeros of $f \left(x\right)$ are expressible in the form $\frac{p}{q}$ for integers $p , q$ with $p$ a divisor of the constant term $- 15$ and $q$ a divisor of the coefficient $16$ of the leading term.

That means that the only possible rational zeros are:

$\pm \frac{1}{16} , \pm \frac{1}{8} , \pm \frac{3}{16} , \pm \frac{1}{4} , \pm \frac{5}{16} , \pm \frac{3}{8} , \pm \frac{1}{2} , \pm \frac{5}{8} , \pm \frac{3}{4} , \pm \frac{15}{16} , \pm 1 , \pm \frac{5}{4} , \pm \frac{3}{2} , \pm \frac{15}{8} , \pm \frac{5}{3} , \pm 3 , \pm 5 , \pm 15$

That is rather a lot of possibilities to try, but we can cut it down radically as follows:

Note that:

$f \left(x\right) = 16 {x}^{4} + 8 {x}^{3} + 4 {x}^{2} - 2 x - 15 = {\left(2 x\right)}^{4} + {\left(2 x\right)}^{3} + {\left(2 x\right)}^{2} - \left(2 x\right) - 15$

So the only possible rational values of $2 x$ making $f \left(x\right) = 0$ are $\pm 1 , \pm 3 , \pm 5 , \pm 15$. Hence the only possible rational zeros are:

$\pm \frac{1}{2} , \pm \frac{3}{2} , \pm \frac{5}{2} , \pm \frac{15}{2}$

None of these work. So $f \left(x\right)$ has no rational zeros.

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Descartes' Rule of Signs

The coefficients of $f \left(x\right)$ have signs in the pattern $+ + + - -$. With one change of sign, that means that $f \left(x\right)$ has exactly $1$ positive Real zero.

The coefficients of $f \left(- x\right)$ have signs in the pattern $+ - + + -$. With $3$ changes of sign, that means that $f \left(x\right)$ has exactly $1$ or $3$ negative Real zeros.

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Where do we go from here?

It is possible to solve $f \left(x\right) = 0$ algebraically, but it gets very long and messy. See https://socratic.org/s/axciCVkR for a typical quartic. Alternatively you can use a numerical method such as Durand-Kerner to find approximations:

${x}_{1} \approx - 1.0194$

${x}_{2} \approx 0.85875$

${x}_{3 , 4} \approx - 0.16967 \pm 1.02084 i$