How do you find all zeros with multiplicities of f(x)=3x^3+3x^2-11x-10?

Apr 10, 2017

The zeros are $- 2$, $\frac{1}{2} + \frac{\sqrt{69}}{6}$, $\frac{1}{2} - \frac{\sqrt{69}}{6}$ all with multiplicity $1$.

Explanation:

Given:

$f \left(x\right) = 3 {x}^{3} + 3 {x}^{2} - 11 x - 10$

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 $- 10$ and $q$ a divisor of the coefficient $3$ of the leading term.

That means that the only possible rational zeros are:

$\pm \frac{1}{3} , \pm \frac{2}{3} , \pm 1 , \pm \frac{5}{3} , \pm 2 , \pm \frac{10}{3} , \pm 5 , \pm 10$

Trying each in turn, we eventually find:

$f \left(- 2\right) = 3 {\left(\textcolor{b l u e}{- 2}\right)}^{3} + 3 {\left(\textcolor{b l u e}{- 2}\right)}^{2} - 11 \left(\textcolor{b l u e}{- 2}\right) - 10$

$\textcolor{w h i t e}{f \left(- 2\right)} = - 24 + 12 + 22 - 10$

$\textcolor{w h i t e}{f \left(- 2\right)} = 0$

So $- 2$ is a zero and $\left(x + 2\right)$ a factor:

$3 {x}^{3} + 3 {x}^{2} - 11 x - 10 = \left(x + 2\right) \left(3 {x}^{2} - 3 x - 5\right)$

The remaining quadratic factor is in the form:

$a {x}^{2} + b x + c$

with $a = 3$, $b = - 3$ and $c = - 5$

We can find its zeros using the quadratic formula:

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

$\textcolor{w h i t e}{x} = \frac{3 \pm \sqrt{{\left(- 3\right)}^{2} - 4 \left(3\right) \left(- 5\right)}}{2 \cdot 3}$

$\textcolor{w h i t e}{x} = \frac{3 \pm \sqrt{9 + 60}}{2 \cdot 3}$

$\textcolor{w h i t e}{x} = \frac{1}{2} \pm \frac{\sqrt{69}}{6}$