# How do you find all the zeros of f(x)=x^3 -8x^2 -23x +30?

Jun 20, 2016

They are $x = 1 , x = 10 , x = - 3$.

#### Explanation:

To solve a cubic equation it exists the solving formula but it is very long and I do not know it.
Then I use a bit of luck to solve this equation. For example I see that $x = 1$ is a solution because

${1}^{3} - 8 \cdot {1}^{2} - 23 \cdot 2 + 30 = 1 - 8 - 23 + 30 = 0$.

Then I know that the equation has to be in the form

$\left(a {x}^{2} + b x + c\right) \left(x - 1\right)$

To find $a , b , c$ I just do the multiplication and compare the coefficients.

$\left(a {x}^{2} + b x + c\right) \left(x - 1\right)$

$= a {x}^{3} + b {x}^{2} + c x - a {x}^{2} - b x - c$

$= a {x}^{3} + \left(b - a\right) {x}^{2} + \left(c - b\right) x - c$

Comparing this with our initial equation we see that

$a = 1$
$b - a = - 8 , b - 1 = - 8 , b = - 7$
$c - b = - 23 , c + 7 = - 23 , c = - 30$
$- c = 30 , c = - 30$

Our function is then

$\left({x}^{2} - 7 x - 30\right) \left(x - 1\right)$.

Now we have to solve a second order equation $\left({x}^{2} - 7 x - 30\right)$ using the solution

$x = \frac{- b \setminus \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

$= \frac{- \left(- 7\right) \setminus \pm \sqrt{{\left(- 7\right)}^{2} - 4 \cdot 1 \cdot \left(- 30\right)}}{2}$

$= \frac{7 \setminus \pm \sqrt{49 + 120}}{2}$

$= \frac{7 \setminus \pm \sqrt{169}}{2}$

$= \frac{7 \setminus \pm 13}{2}$

that has the two solutions

$x = \frac{7 + 13}{2} = 10$ and $x = \frac{7 - 13}{2} = - 3$.

Then the function

$f \left(x\right) = {x}^{3} - 8 {x}^{2} - 23 x + 30$

can be rewritten as

$f \left(x\right) = \left(x - 1\right) \left(x - 10\right) \left(x + 3\right)$ with the three zeros for $x$ equal to $1 , 10 , - 3$.