# How do you find all the zeros of x^3-7x-6?

May 23, 2016

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

#### Explanation:

Given a polynomial in which the maximum power therm coefficient is 1, the constant therm is the product of its roots.

Examining the polynomial

${p}_{3} \left(x\right) = {x}^{3} - 7 x - 6$

we can conclude that $6 = {x}_{1} \cdot {x}_{2} \cdot {x}_{3}$
such that ${p}_{3} \left(x\right) = \left(x - {x}_{1}\right) \left(x - {x}_{2}\right) \left(x - {x}_{3}\right)$.

Supposing that the roots are integers we can try the set of values

$\left\{\pm 1 , \pm 2 , \pm 3\right\}$ which are potential $- 6$ factors

Easily we can verify that

${p}_{3} \left(- 1\right) = {p}_{3} \left(- 2\right) = {p}_{3} \left(3\right) = 0$ so we found the three roots and we can state:

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