# How do you find all the zeros of f(x)=3x^4-17x^3+33x^2-17x-10?

Apr 16, 2016

Find zeros: $- \frac{1}{3}$, $2$ and $2 \pm i$

#### Explanation:

By the rational root theorem, any rational zeros of $f \left(x\right)$ must be expressible in the form $\frac{p}{q}$ for integers $p , q$ 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 find:

$f \left(- \frac{1}{3}\right) = \frac{3}{81} + \frac{17}{27} + \frac{33}{9} + \frac{17}{3} - 10$

$= \frac{1 + 17 + 99 + 153 - 270}{27} = 0$

$f \left(2\right) = 48 - 136 + 132 - 34 + 10 = 0$

So $x = - \frac{1}{3}$ and $x = 2$ are zeros and $\left(3 x + 1\right)$ and $\left(x - 2\right)$ are factors of $f \left(x\right)$:

$3 {x}^{4} - 17 {x}^{3} + 33 {x}^{2} - 17 x - 10$

$= \left(3 x + 1\right) \left({x}^{3} - 6 {x}^{2} + 13 x - 10\right)$

$= \left(3 x + 1\right) \left(x - 2\right) \left({x}^{2} - 4 x + 5\right)$

The remaining quadratic factor has negative discriminant, but we can factor it using Complex numbers, completeing the square and the difference of squares identity:

${a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)$

as follows:

${x}^{2} - 4 x + 5 = {\left(x - 2\right)}^{2} + 1 = {\left(x - 2\right)}^{2} - {i}^{2} = \left(\left(x - 2\right) - i\right) \left(\left(x - 2\right) + i\right) = \left(x - 2 - i\right) \left(x - 2 + i\right)$

So the remaining zeros are $x = 2 \pm i$