# How do you find all the real and complex roots and use Descartes Rule of Signs to analyze the zeros of P(x) = 12x^4 + x^3 + 4x^2 + 7x + 8 ?

Aug 2, 2016

This quartic has $4$ Complex non-Real zeros.

#### Explanation:

$P \left(x\right) = 12 {x}^{4} + {x}^{3} + 4 {x}^{2} + 7 x + 8$

By the rational root theorem, any rational zeros of $P \left(x\right)$ are expressible in the form $\frac{p}{q}$ for integers $p , q$ with $p$ a divisor of the constant term $8$ and $q$ a divisor of the coefficient $12$ of the leading term.

That means that the only possible rational zeros are:

$\pm \frac{1}{12} , \pm \frac{1}{6} , \pm \frac{1}{4} , \pm \frac{1}{3} , + \frac{1}{2} , \pm \frac{2}{3} , \pm 1 , \pm \frac{4}{3} , \pm 2 , \pm \frac{8}{3} , \pm 4 , \pm 8$

In addition note that all of the signs of the coefficients are positive, so there are no changes of sign. So by Descartes rule of signs there are no positive zeros.

Reversing the signs on the terms with odd degree, we get the pattern of signs: $+ - + - +$. With $4$ changes of sign, that means that there are $4$, $2$ or $0$ negative Real zeros.

So the only possible rational zeros are:

$- \frac{1}{12} , - \frac{1}{6} , - \frac{1}{4} , - \frac{1}{3} , - \frac{1}{2} , - \frac{2}{3} , - 1 , - \frac{4}{3} , - 2 , - \frac{8}{3} , - 4 , - 8$

That leaves $0$, $2$ or $4$ non-Real Complex zeros, occuring in Complex conjugate pairs.

Trying each of the possible rational zeros in turn, we find that $P \left(x\right) > 0$ for all of them. So $P \left(x\right)$ has no rational zeros.

Though it is possible to find the zeros analytically, it gets very messy. It is substantially easier to find approximations to the zeros using a numerical method such as Durand-Kerner to find:

${x}_{1 , 2} \approx 0.56217 \pm 0.86572 i$

${x}_{3 , 4} \approx - 0.60384 \pm 0.51093 i$