# How do you find the zeros of  x^3 - 7x + 6 = 0?

Jun 14, 2016

${x}^{3} - 7 x + 6 = \left(x - 1\right) \left(x - 2\right) \left(x + 3\right)$

#### Explanation:

We use the property that if $\alpha$ is zero of $f \left(x\right)$ i.e. $f \left(\alpha\right) = 0$, then $\left(x - \alpha\right)$ is a factor of $f \left(x\right)$.

As here $f \left(x\right) = {x}^{3} - 7 x + 6$ and $f \left(1\right) = {1}^{3} - 7 \cdot 1 + 6 = 1 - 7 + 6 = 0$

$\left(x - 1\right)$ is a factor of ${x}^{3} - 7 x + 6$.

Now dividing ${x}^{3} - 7 x + 6$ by $\left(x - 1\right)$, we get

${x}^{2} + x - 6$, whose discriminant is ${1}^{2} - 4 \cdot 1 \cdot \left(- 6\right) = 25 = {5}^{2}$, hence factors can be found by splitting middle term. Hence,

${x}^{2} + x - 6 = {x}^{2} + 3 x - 2 x - 6 = x \left(x + 3\right) - 2 \left(x + 3\right) = \left(x - 2\right) \left(x + 3\right)$

Hence, ${x}^{3} - 7 x + 6 = \left(x - 1\right) \left(x - 2\right) \left(x + 3\right)$