# How do you find the zeroes of p(x)= x^3+2x^2-9x-8?

Nov 8, 2015

Try rational root theorem, but find there are no rational roots.

Use general cubic formula.

#### Explanation:

By the rational root theorem any rational roots of $p \left(x\right) = 0$ must be divisors of $8$. So the only possible rational roots are $\pm 1$, $\pm 2$, $\pm 4$, $\pm 8$.

We find:

$p \left(- 4\right) = - 4$
$p \left(- 2\right) = 10$
$p \left(- 1\right) = 2$
$p \left(1\right) = - 14$
$p \left(2\right) = - 10$
$p \left(4\right) = 52$

So the three Real roots of $p \left(x\right) = 0$ lie in the intervals $\left(- 4 , - 2\right)$, $\left(- 1 , 1\right)$ and $\left(2 , 4\right)$ and there are no rational roots.

Here's an algebraic solution using the general cubic formula:

${x}^{3} + 2 {x}^{2} - 9 x - 8$ is of the form $a {x}^{3} + b {x}^{2} + c x + d$ with $a = 1$, $b = 2$, $c = - 9$ and $d = - 8$.

This has resolvents ${\Delta}_{0}$, ${\Delta}_{1}$, $C$ given by the following formulae:

${\Delta}_{0} = {b}^{2} - 3 a c = {2}^{2} - \left(3 \cdot 1 \cdot - 9\right) = 4 + 27 = 31$

${\Delta}_{1} = 2 {b}^{3} - 9 a b c + 27 {a}^{2} d = \left(2 \cdot {2}^{3}\right) - \left(9 \cdot 1 \cdot 2 \cdot - 9\right) + \left(27 \cdot {1}^{2} \cdot - 8\right) = 16 + 162 - 216 = - 38$

$C = \sqrt[3]{\frac{{\Delta}_{1} + \sqrt{{\Delta}_{1}^{2} - 4 {\Delta}_{0}^{3}}}{2}}$

$= \sqrt[3]{\frac{- 38 + \sqrt{{38}^{2} - 4 \cdot {31}^{3}}}{2}}$

$= \sqrt[3]{- 19 \pm \sqrt{{19}^{2} - {31}^{3}}}$

$= \sqrt[3]{- 19 \pm \sqrt{361 - 29791}}$

$= \sqrt[3]{- 19 \pm \sqrt{- 29430}}$

$= \sqrt[3]{- 19 \pm 3 \sqrt{3270} i}$

Then the roots are:

${x}_{n} = - \frac{1}{3 a} \left(b + {\omega}^{n} C + {\Delta}_{0} / \left({\omega}^{n} C\right)\right)$

$= - \frac{1}{3} \left(2 + {\omega}^{n} \sqrt[3]{- 19 \pm 3 \sqrt{3270} i} + \frac{31}{{\omega}^{n} \sqrt[3]{- 19 \pm 3 \sqrt{3270} i}}\right)$

where $\omega = \frac{- 1 + \sqrt{3} i}{2}$ and $n = 0 , 1 , 2$

Nov 8, 2015

Find the roots numerically using Newton's method.

${x}_{1} \approx - 3.810821$
${x}_{2} \approx - 0.803113$
${x}_{3} \approx 2.613934$

#### Explanation:

When seeking a rational root we find:

$p \left(- 4\right) = - 4$
$p \left(- 2\right) = 10$
$p \left(- 1\right) = 2$
$p \left(1\right) = - 14$
$p \left(2\right) = - 10$
$p \left(4\right) = 52$

Hence $p \left(x\right) = 0$ has irrational Real roots in $\left(- 4 , - 2\right)$, $\left(- 1 , 1\right)$ and $\left(2 , 4\right)$.

$p \left(x\right) = {x}^{3} + 2 {x}^{2} - 9 x - 8$

$p ' \left(x\right) = 3 {x}^{2} + 4 x - 9$

Newton's method starts with an approximation ${a}_{0}$ for the root, then applies a formula to refine that approximation iteratively:

${a}_{i + 1} = {a}_{i} - \frac{p \left({a}_{i}\right)}{p ' \left({a}_{i}\right)}$

$= {a}_{i} - \frac{{a}_{i}^{3} + 2 {a}_{i}^{2} - 9 {a}_{i} - 8}{3 {a}_{i}^{2} + 4 {a}_{i} - 9}$

Putting this formula into a spreadsheet and using initial approximations ${a}_{0} = - 4$, ${a}_{0} = - 1$, ${a}_{0} = 2$ we get the following sequences of approximations:

${a}_{0} = - 4$
${a}_{1} = - 3.826087$
${a}_{2} = - 3.810933$
${a}_{3} = - 3.810821$

${a}_{0} = - 1$
${a}_{1} = - 0.8$
${a}_{2} = - 0.803113$

${a}_{0} = 2$
${a}_{1} = 2.909091$
${a}_{2} = 2.646363$
${a}_{3} = 2.613934$