# How do you find all the real and complex roots of  P(x) = x^5 + 3x^3 + 2x - 6?

Feb 21, 2016

Here's a partial answer. I may provide another (numerical) approach later...

#### Explanation:

First note that the sum of the coefficients is zero.

That is: $1 + 3 + 2 - 6 = 0$

So $P \left(1\right) = 0$ and $\left(x - 1\right)$ is a factor:

${x}^{5} + 3 {x}^{3} + 2 x - 6 = \left(x - 1\right) \left({x}^{4} + {x}^{3} + 4 {x}^{2} + 4 x + 6\right)$

By the rational root theorem any rational zeros of $f \left(x\right) = {x}^{4} + {x}^{3} + 4 {x}^{2} + 4 x + 6$ must be expressible in the form $\frac{p}{q}$ for integers $p$ and $q$ such that $p$ is a divisor of the constant term $6$ and $q$ a divisor of the coefficient $1$ of the leading term.

In addition, all of the coefficients of $f \left(x\right)$ are positive, so it has no positive zeros. so the only possible rational zeros are:

$- 1$, $- 2$, $- 3$, $- 6$

None of these work, so $f \left(x\right) = 0$ has no rational roots.

It is possible to solve $f \left(x\right) = 0$ algebraically, but it gets rather messy.

Here's a quick sketch of how to do it:

First multiply $f \left(x\right)$ by ${4}^{4} = 256$ and substitute $t = \left(4 x + 1\right)$ to get the equation:

${t}^{4} + 58 {t}^{2} + 139 t + 1338 = 0$

Since this has no cubic term, it has a factorisation of the form:

${t}^{4} + 58 {t}^{2} + 139 t + 1338 = \left({t}^{2} + A t + B\right) \left({t}^{2} - A t + C\right)$

Multiply this out and equate coefficients to get three simultaneous equations in $A$, $B$ and $C$.

Then use ${\left(B + C\right)}^{2} = {\left(C - B\right)}^{2} + 4 B C$ to derive a cubic equation in ${A}^{2}$:

${\left({A}^{2}\right)}^{3} + 116 {\left({A}^{2}\right)}^{2} - 1988 \left({A}^{2}\right) - 19321 = 0$

To solve this, multiply by $27$ and substitute $s = 3 {A}^{2} + 116$ to get a cubic equation in $s$:

${s}^{3} - 58260 s + 4675597 = 0$

This cubic has $3$ Real roots, and none of the Real roots are expressible in $a + b i$ form with $a$ and $b$ expressed in terms of Real square and cube roots.

So we have the choice of expressing them in terms of Complex cube roots (using Cardano's method or Vieta's substitution) or using the trigonometric cubic solution, expressing the roots in terms of $\cos$ and $\arccos$.

If we have expressions for the roots of our cubic in $s$, then we can deduce expressions for ${A}^{2}$, hence $B$ and $C$, providing quadratic equations that we can solve using the quadratic formula.

This is way too messy, so I would recommend using a numeric method such as Newton's instead.