# How do you find all the zeros of f(x) = x^3 - 4x^(2) + 14x - 20?

Jun 5, 2016

$x = 2$ or $x = 1 \pm 3 i$

#### Explanation:

$f \left(x\right) = {x}^{3} - 4 {x}^{2} + 14 x - 20$

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 $- 20$ and $q$ a divisor of the coefficient $1$ of the leading term.

That means that the only possible rational zeros are:

$\pm 1$, $\pm 2$, $\pm 4$, $\pm 5$, $\pm 10$, $\pm 20$

In addition note that if you invert the signs of the coefficients of the terms of odd degree, then the pattern of signs you get is:

$- - - -$

which has no changes. So there are no negative zeros.

So the only possible rational zeros are:

$1 , 2 , 4 , 5 , 10 , 20$

We find:

$f \left(2\right) = 8 - 16 + 28 - 20 = 0$

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

${x}^{3} - 4 {x}^{2} + 14 x - 20 = \left(x - 2\right) \left({x}^{2} - 2 x + 10\right)$

The remaining quadratic factor has negative discriminant, so no Real zeros, but we can still factor it by completing the square and using the difference of squares identity:

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

with $a = \left(x - 1\right)$ and $b = 3 i$ as follows:

${x}^{2} - 2 x + 10$

$= {x}^{2} - 2 x + 1 + {3}^{2}$

$= {\left(x - 1\right)}^{2} - {\left(3 i\right)}^{2}$

$= \left(\left(x - 1\right) - 3 i\right) \left(\left(x - 1\right) + 3 i\right)$

$= \left(x - 1 - 3 i\right) \left(x - 1 + 3 i\right)$

Hence Complex zeros:

$x = 1 \pm 3 i$