# How do you find all the zeros of f(x) = x^3 - 3x^2 -15x +125?

Feb 25, 2016

Three zeros are real $x = - 5$, and imaginary $x = 4 \pm 3 i$

#### Explanation:

We know that if $\left(x + a\right)$ is a zero of a function $f \left(x\right)$ then $f \left(- a\right) = 0$.
Given is $f \left(x\right) = {x}^{3} - 3 {x}^{2} - 15 x + 125$

Let us choose one factor as $\left(x + 5\right)$. This is trial and error, notice that there is $125$ as the last term therefore this is a possible factor.
Now $f \left(- 5\right) = {\left(- 5\right)}^{3} - 3 {\left(- 5\right)}^{2} - 15 \left(- 5\right) + 125$
or $f \left(- 5\right) = - 125 - 75 + 75 + 125 = 0$
Hence $\left(x + 5\right)$ is a factor.
Writing the given equation as a multiplication of the factor and a quadratic in general form

$\left(x + 5\right) \left(a {x}^{2} + b x + c\right) = {x}^{3} - 3 {x}^{2} - 15 x + 125$

$a {x}^{3} + b {x}^{2} + c x + 5 a {x}^{2} + 5 b x + 5 c = {x}^{3} - 3 {x}^{2} - 15 x + 125$

Put like terms together

$a {x}^{3} + b {x}^{2} + 5 a {x}^{2} + c x + 5 b x + 5 c = {x}^{3} - 3 {x}^{2} - 15 x + 125$

$a {x}^{3} + {x}^{2} \left(b + 5 a\right) + x \left(c + 5 b\right) + 5 c = {x}^{3} - 3 {x}^{2} - 15 x + 125$

Compare coefficients of like terms
$a = 1$
$b + 5 a = - 3$ , Inserting value of $a$ gives us $b = - 8$
$c + 5 b = - 15$, Inserting value of $b$ gives us $c = 25$
$5 c = 125$ , Gives us again $c = 25$

Now rewrite polynomial as
$f \left(x\right) = \left(x + 5\right) \left({x}^{2} - 8 x + 25\right)$

To find remaining two zeros using quadratic formula we obtain

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

$x = \frac{- \left(- 8\right) \pm \sqrt{{\left(- 8\right)}^{2} - 4 \times 1 \times 25}}{2 \times 1}$

$x = \frac{8 \pm \sqrt{64 - 100}}{2}$
$x = \frac{8 \pm \sqrt{- 36}}{2}$
$x = \frac{8 \pm 6 i}{2}$
$x = 4 \pm 3 i$