# How do you find all the real and complex roots and use Descartes Rule of Signs to analyze the zeros of  P(x) = 2x^5 + 7x^3 + 6x^2 - 2 ?

Jul 31, 2016

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#### Explanation:

$P \left(x\right) = 2 {x}^{5} + 7 {x}^{3} + 6 {x}^{2} - 2$

The pattern of signs of the coefficients is $+ + + -$. Since there is only one change of signs, this quintic has exactly $1$ positive Real zero.

$P \left(- x\right) = - 2 {x}^{5} - 7 {x}^{3} + 6 {x}^{2} - 2$

The pattern of signs of the coefficients of $P \left(- x\right)$ is $- - + -$. Since there are $2$ changes of signs, $P \left(x\right)$ has exactly $0$ or $2$ negative Real zeros.

By the fundamental theorem of algebra, $P \left(x\right)$ has exactly $5$ zeros counting multiplicity, so that leaves $4$ or $2$ Complex zeros, occuring in Complex conjugate pairs.

By the rational root theorem, any rational zeros of $P \left(x\right)$ must be expressible in the form $\frac{p}{q}$ for integers $p , q$ with $p$ a divisor of the constant term $- 2$ and $q$ a divisor of the coefficient $2$ of the leading term.

That means that the only possible rational zeros are:

$\pm \frac{1}{2} , \pm 1 , \pm 2$

None of these work, so $P \left(x\right)$ has no rational zeros.

In common with most quintics and higher order polynomials, the zeros of $P \left(x\right)$ are not expressible using elementary functions, including $n$th roots and trigonometric functions.

About the best you can do is use a numerical method such as Durand-Kerner to find approximations for the zeros:

${x}_{1} \approx 0.460754$

${x}_{2 , 3} \approx - 0.617273 \pm 0.377799 i$

${x}_{4 , 5} \approx 0.386896 \pm 1.99853 i$