# How do you find all the zeros of f(x) = x^3 + 11x^2 + 41x + 51?

Mar 21, 2016

Use the rational root theorem to find zero $x = - 3$, divide by $\left(x + 3\right)$ then solve the remaining quadratic to find zeros $x = - 4 \pm i$.

#### Explanation:

$f \left(x\right) = {x}^{3} + 11 {x}^{2} + 41 x + 51$

by the rational root theorem, any rational zeros of $f \left(x\right)$ must be expressible in the form $\frac{p}{q}$ for integers $p$ and $q$ where $p$ is a divisor of the constant term $51$ and $q$ a divisor of the coefficient $1$ of the leading term.

That means that the only possible rational roots are:

$\pm 1$, $\pm 3$, $\pm 17$, $\pm 51$

In addition, note that all of the coefficients of $f \left(x\right)$ are positive, so it can only have zeros for negative values of $x$.

That leaves:

$- 1$, $- 3$, $- 17$, $- 51$

Trying each in turn we find:

$f \left(- 1\right) = - 1 + 11 - 41 + 51 = 20$

$f \left(- 3\right) = - 27 + 99 - 123 + 51 = 0$

So $x = - 3$ is a zero and $\left(x + 3\right)$ a factor:

${x}^{3} + 11 {x}^{2} + 41 x + 51 = \left(x + 3\right) \left({x}^{2} + 8 x + 17\right)$

Then solve:

$0 = {x}^{2} + 8 x + 17$

$= {x}^{2} + 8 x + 16 + 1$

$= {\left(x + 4\right)}^{2} - {i}^{2}$

$= \left(\left(x + 4\right) - i\right) \left(\left(x + 4\right) + i\right)$

$= \left(x + 4 - i\right) \left(x + 4 + i\right)$

So the remaining zeros are: $x = - 4 \pm i$