# How do you find all the zeros of f(x)=x^3 + 4x^2 - 5?

${x}_{1} = 1$ and ${x}_{2} = \frac{- 5 + \sqrt{5}}{2}$ and ${x}_{3} = \frac{- 5 - \sqrt{5}}{2}$

#### Explanation:

From the given:
$y = {x}^{3} + 4 {x}^{2} - 5$

Perform synthetic division

${x}^{3} \text{ " " "x^2" " " "x^1" " " " } {x}^{0}$

$1 \text{ " " " " 4" " " " "0" " } - 5$ trial divisor $= 1$
$\underline{\text{ " " " " " "1" " " " " "5" " " " " } 5}$
$1 \text{ " " " " 5" " " " " "5" " " " " } 0$$\leftarrow$remainder

${x}_{1} = 1$ is a zero

depressed equation:

${x}^{2} + 5 x + 5 = 0$

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

Let $a = 1$ and $b = 5$ and $c = 5$

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

$x = \frac{- 5 \pm \sqrt{{5}^{2} - 4 \left(1\right) \left(5\right)}}{2 \left(1\right)}$

$x = \frac{- 5 \pm \sqrt{5}}{2}$

the other zeros are:
${x}_{2} = \frac{- 5 + \sqrt{5}}{2}$

${x}_{3} = \frac{- 5 - \sqrt{5}}{2}$

God bless....I hope the explanation is useful.