# How do you find the number of complex, real and rational roots of 4x^5-5x^4+x^3-2x^2+2x-6=0?

Jan 11, 2017

This quintic equation has one positive irrational Real zero and two complex conjugate pairs of non-Real Complex zeros.

#### Explanation:

Given:

$f \left(x\right) = 4 {x}^{5} - 5 {x}^{4} + {x}^{3} - 2 {x}^{2} + 2 x - 6$

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Fundamental Theorem of Algebra

The Fundamental Theorem of Algebra (FTOA) tells us that a polynomial in one variable of degree $n > 0$ has a Complex (possibly Real) zero.

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

In our example, $f \left(x\right)$ is of degree $5$, so has exactly $5$ Complex (possibly Real) zeros.

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Descartes' Rule of Signs

The signs of $f \left(x\right)$ are in the pattern: $+ - + - + -$. With $5$ changes of sign, Descartes' Rule of Signs tells us that $f \left(x\right)$ has $5$, $3$ or $1$ positive Real zero.

The signs of $f \left(- x\right)$ are in the pattern $- - - - - -$. With $0$ changes of sign, we can deduce that $f \left(x\right)$ has no negative Real zeros.

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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 $- 6$ and $q$ a divisor of the coefficient $4$ of the leading term.

In addition, from Descartes' Rule of Signs, we know that such zeros can only be positive. So the only possible rational zeros are:

$\frac{1}{4} , \frac{1}{2} , \frac{3}{4} , 1 , \frac{3}{2} , 2 , 3 , 6$

We find:

$f \left(\frac{1}{4}\right) = - \frac{45}{8}$

$f \left(\frac{1}{2}\right) = - \frac{89}{16}$

$f \left(\frac{3}{4}\right) = - \frac{747}{128}$

$f \left(1\right) = - 6$

$f \left(\frac{3}{2}\right) = \frac{15}{16}$

$f \left(2\right) = 46$

$f \left(3\right) = 576$

$f \left(6\right) = 24774$

So $f \left(x\right)$ has no rational zeros. It has an irrational Real zero somewhere in $\left(1 , \frac{3}{2}\right)$

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Durand-Kerner Algorithm

We can use the Durand-Kerner algorithm to find numerical approximations to the zeros.

In our example, we find approximations:

${x}_{1} \approx 1.47294$

${x}_{2 , 3} \approx 0.547393 \pm 0.873706 i$

${x}_{4 , 5} \approx - 0.658864 \pm 0.723817 i$

So we find that $f \left(x\right)$ has one Real and four non-Real Complex zeros.