# What are all the possible rational zeros for f(x)=x^3-4x^2-3x+14 and how do you find all zeros?

Aug 18, 2016

$f \left(x\right)$ has zeros $2$ and $1 \pm 2 \sqrt{2}$

#### Explanation:

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

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 of the constant term $14$ and $q$ a divisor of the coefficient $1$ of the leading term.

That means that the only possible rational zeros are:

$\pm 1 , \pm 2 , \pm 7 , \pm 14$

Trying each in turn, we find:

$f \left(2\right) = 8 - 4 \left(4\right) - 3 \left(2\right) + 14 = 8 - 16 - 6 + 14 = 0$

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

${x}^{3} - 4 {x}^{2} - 3 x + 14 = \left(x - 2\right) \left({x}^{2} - 2 x - 7\right)$

We can factor the remaining quadratic by completing the square:

${x}^{2} - 2 x - 7$

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

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

$= \left(\left(x - 1\right) - 2 \sqrt{2}\right) \left(\left(x - 1\right) + 2 \sqrt{2}\right)$

=(x-(1+2sqrt(2))(x-(1-2sqrt(2)))

Hence zeros:

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