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

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Answer 1 · 2016-05-23T14:19:26

$x^3-7x-6=(x+1)(x+2)(x-3)$

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

Examining the polynomial

$p_3(x) = x^3-7x-6$

we can conclude that $6 = x_1*x_2*x_3$
such that $p_3(x) = (x-x_1)(x-x_2)(x-x_3)$ .

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

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

Easily we can verify that

$p_3(-1) = p_3(-2) = p_3(3) = 0$ so we found the three roots and we can state:

$x^3-7x-6=(x+1)(x+2)(x-3)$

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