How do you find all the zeros of f(x)=x^3-2x^2-11x+52?

Aug 4, 2016

$f \left(x\right)$ has zeros $- 4$ and $3 \pm 2 i$

Explanation:

$f \left(x\right) = {x}^{3} - 2 {x}^{2} - 11 x + 52$

By the rational root 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 $52$ and $q$ a divisor of the coefficient $1$ of the leading term.

That means that the only possible rational zeros are:

$\pm 1 , \pm 2 , \pm 4 , \pm 13 , \pm 26 , \pm 52$

In addition, note that the signs of the coefficients are in the pattern $+ - - +$, having $2$ changes of sign, so by Descartes' rule of signs, $f \left(x\right)$ has $2$ or $0$ positive Real zeros.

The pattern of signs of coefficients of $f \left(- x\right)$ is $- - + +$. With just one change of sign, $f \left(x\right)$ must have exactly one negative Real zero.

So check for negative rational zeros first:

$f \left(- 1\right) = - 1 - 2 + 11 + 52 = 60$

$f \left(- 2\right) = - 8 - 8 + 22 + 52 = 58$

$f \left(- 4\right) = - 64 - 32 + 44 + 52 = 0$

So $x = - 4$ is a zero and $\left(x + 4\right)$ a factor:

${x}^{3} - 2 {x}^{2} - 11 x + 52 = \left(x + 4\right) \left({x}^{2} - 6 x + 13\right)$

The discriminant of the remaining quadratic factor is negative, so it has no Real zeros, but we can factor it with Complex coefficients:

${x}^{2} - 6 x + 13$

$= {\left(x - 3\right)}^{2} - 9 + 13$

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

$= {\left(x - 3\right)}^{2} - {\left(2 i\right)}^{2}$

$= \left(x - 3 - 2 i\right) \left(x - 3 + 2 i\right)$

Hence zeros: $x = 3 \pm 2 i$