Answer:
{-2, -1, 3}
Explanation:
You want the zeros of F(x)=x^3-7x-6.
Descartes' rule of signs
The signs of the coefficients in order of decreasing degree are +--. There is one sign change, so there is one positive real root.
Rational root theorem
The rational root theorem tells you any real roots will be divisors of the constant, 6. Hence, we only need to try ±1, ±2, ±3.
The sum of coefficients is 1-7-6 = -12, so x=1 is not a solution. The positive solution is greater than +1.
The sum of coefficients of odd-degree terms is (1 -7) = -6, equal to the sum of coefficients of even-degree terms (0 -6) = -6, so -1 is a root.
Factorization
The attachment shows F(x) divided by the corresponding factor (x+1) using synthetic division. The quotient is (x² -x -6), so we know the factorization is ...
F(x) = (x +1)(x² -x -6) = (x +1)(x -3)(x +2)
The zero product rule tells us the roots are ...
x = {-2, -1, 3}
The graph of the function tells us the same thing.
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Additional comment
The constants in the binomial factors of x² -x -6 will be factors of the constant term, -6, that sum to the linear term coefficient, -1. These are -3 and +2.
We like a graphing calculator for finding roots of a cubic and higher-degree polynomials. It will show only the real roots, so more steps are required for complex roots.