What are all the possible rational zeros for y=x^3-5x^2+17x-13 and how do you find all zeros?

Sep 5, 2016

The zeros are $x = 1$ and $x = 2 \pm 3 i$

Explanation:

$y = {x}^{3} - 5 {x}^{2} + 17 x - 13$

By the rational roots theorem, any rational zeros of this cubic are expressible in the form $\frac{p}{q}$ for integers $p , q$ with $p$ a divisor of the constant term $- 13$ and $q$ a divisor of the coefficient $1$ of the leading term.

In addition note that there are no negative zeros, since reversing the signs of the coefficients of the terms of odd degree results in all positive coefficients.

Hence the only possible rational zeros are:

$1 , 13$

Also note that the sum of the coefficients is zero, so $x = 1$ is a zero and $\left(x - 1\right)$ a factor:

${x}^{3} - 5 {x}^{2} + 17 x - 13 = \left(x - 1\right) \left({x}^{2} - 4 x + 13\right)$

We can find the zeros of the remaining quadratic by completing the square and using the difference of squares identity:

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

with $a = \left(x - 2\right)$ and $b = 3 i$ as follows:

$0 = {x}^{2} - 4 x + 13$

$\textcolor{w h i t e}{0} = {x}^{2} - 4 x + 4 + 9$

$\textcolor{w h i t e}{0} = {\left(x - 2\right)}^{2} - {\left(3 i\right)}^{2}$

$\textcolor{w h i t e}{0} = \left(\left(x - 2\right) - 3 i\right) \left(\left(x - 2\right) + 3 i\right)$

$\textcolor{w h i t e}{0} = \left(x - 2 - 3 i\right) \left(x - 2 + 3 i\right)$

So the remaining two zeros are $x = 2 \pm 3 i$