# What are all the possible rational zeros for f(x)=3x^3-7x^2+29x-9 and how do you find all zeros?

Dec 31, 2016

The "possible" rational zeros are:

$\pm \frac{1}{3} , \pm 1 , \pm 3 , \pm 9$

The actual zeros are:

$\frac{1}{3}$ and $1 \pm 2 \sqrt{2} i$

#### Explanation:

$f \left(x\right) = 3 {x}^{3} - 7 {x}^{2} + 29 x - 9$

By the rational roots 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 fo the constant term $- 9$ and $q$ a divisor of the coefficient $3$ of the leading term.

That means that the only possible rational zeros are:

$\pm \frac{1}{3} , \pm 1 , \pm 3 , \pm 9$

Note also that the signs of the coefficients of $f \left(x\right)$ are in the pattern: $+ - + -$. By Descartes's rule of signs, since this has $3$ changes of sign, $f \left(x\right)$ must have $3$ or $1$ positive real zero. Note that $f \left(- x\right) = - 3 {x}^{3} - 7 {x}^{2} - 29 x - 9$ has all negative coefficients, so $f \left(x\right)$ has no negative real zeros.

So the only possible rational real zeros are:

$\frac{1}{3} , 1 , 3 , 9$

Trying each in turn we immediately find:

$f \left(\frac{1}{3}\right) = 3 \left(\frac{1}{27}\right) - 7 \left(\frac{1}{9}\right) + 29 \left(\frac{1}{3}\right) - 9 = \frac{1 - 7 + 87 - 81}{9} = 0$

So $x = \frac{1}{3}$ is a zero and $\left(3 x - 1\right)$ a factor:

$3 {x}^{3} - 7 {x}^{2} + 29 x - 9 = \left(3 x - 1\right) \left({x}^{2} - 2 x + 9\right)$

We can factor the remaining quadratic by completing the square, but it does require some Complex coefficients:

${x}^{2} - 2 x + 9 = {x}^{2} - 2 x + 1 + 8$

$\textcolor{w h i t e}{{x}^{2} - 9 x + 9} = {\left(x - 1\right)}^{2} - {\left(2 \sqrt{2} i\right)}^{2}$

$\textcolor{w h i t e}{{x}^{2} - 9 x + 9} = \left(\left(x - 1\right) - 2 \sqrt{2} i\right) \left(\left(x - 1\right) + 2 \sqrt{2} i\right)$

$\textcolor{w h i t e}{{x}^{2} - 9 x + 9} = \left(x - 1 - 2 \sqrt{2} i\right) \left(x - 1 + 2 \sqrt{2} i\right)$

Hence the other two zeros are:

$x = 1 \pm 2 \sqrt{2} i$