How do you find all the zeros of x^3-7x-6x3−7x−6?

1 Answer
Cesareo R.
May 23, 2016

x^3-7x-6=(x+1)(x+2)(x-3)x3−7x−6=(x+1)(x+2)(x−3)

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(x) = x^3-7x-6p3(x)=x3−7x−6

we can conclude that 6 = x_1*x_2*x_36=x1⋅x2⋅x3
such that p_3(x) = (x-x_1)(x-x_2)(x-x_3)p3(x)=(x−x1)(x−x2)(x−x3).

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

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

Easily we can verify that

p_3(-1) = p_3(-2) = p_3(3) = 0p3(−1)=p3(−2)=p3(3)=0 so we found the three roots and we can state:

x^3-7x-6=(x+1)(x+2)(x-3)x3−7x−6=(x+1)(x+2)(x−3)

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