# How do you find all zeros of the function f(x)= x^3 - x^2 - 4x -6 given 3 as a zero?

Feb 24, 2016

Other two zeros of f(x) are $\left(- 1 + i\right)$ and $\left(- 1 - i\right)$.

#### Explanation:

Zeros of the function f(x)=x^3−x^2−4x−6 are those values of $x$ for which $f \left(x\right) = 0$. As the term independent of $x$ is $- 6$, factors of $- 6$ i.e. $\left\{1 , - 1 , 2 , - 2 , 3 , - 3 , 6 , - 6\right\}$ could be among zeros of $f \left(x\right)$.

It is seen that putting $x = 3$ makes $f \left(x\right) = 0$ and hence $\left(x - 3\right)$ isne such factor. Dividing $f \left(x\right)$ by $\left(x - 3\right)$ we get

x^3−x^2−4x−6

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

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

As the determinant (${b}^{2} - 4 a c$ in the function $a {x}^{2} + b x + c$) is equai to ${2}^{2} - 4 \cdot 1 \cdot 2 = - 4$, is a negative number, no further real zeros are there.

Other complex zeros are given by $\frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

i.e. $\frac{- 2 \pm \sqrt{{2}^{2} - 4 \cdot 1 \cdot 2}}{2 \cdot 1}$ or $\frac{- 2 \pm \sqrt{- 4}}{2}$

which simplifies to $\left(- 1 + i\right)$ and $\left(- 1 - i\right)$, which are other two zeros of $f \left(x\right)$.