# What are all the possible rational zeros for f(x)=x^3+5x^2-11x-6 and how do you find all zeros?

Oct 26, 2016

$f \left(x\right)$ has zeros $2$ and $- \frac{7}{2} \pm \frac{\sqrt{37}}{2}$

#### Explanation:

$f \left(x\right) = {x}^{3} + 5 {x}^{2} - 11 x - 6$

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 $6$ and $q$ a divisor of the coefficient $1$ of the leading term.

Thats means that the only possible rational zeros are:

$\pm 1 , \pm 2 , \pm 3 , \pm 6$

In addition, note that the signs of the coefficients of $f \left(x\right)$ are in the pattern: $+ + - -$. With one change of sign, Descartes' Rule of Signs tells us that $f \left(x\right)$ has exactly one positive Real zero. So let's try the positive rational possibilities first:

$f \left(1\right) = 1 + 5 - 11 + 6 = 1$

$f \left(2\right) = 8 + 5 \left(4\right) - 11 \left(2\right) - 6 = 8 + 20 - 22 - 6 = 0$

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

${x}^{3} + 5 {x}^{2} - 11 x - 6 = \left(x - 2\right) \left({x}^{2} + 7 x + 3\right)$

The remaining two zeros are roots of the quadratic: ${x}^{2} + 7 x + 3 = 0$, which we can solve using the quadratic formula:

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

$\textcolor{w h i t e}{x} = \frac{- 7 \pm \sqrt{{7}^{2} - 4 \left(1\right) \left(3\right)}}{2 \cdot 1}$

$\textcolor{w h i t e}{x} = \frac{- 7 \pm \sqrt{49 - 12}}{2}$

$\textcolor{w h i t e}{x} = - \frac{7}{2} \pm \frac{\sqrt{37}}{2}$