1. **State the problem:** Find the possible real roots of the polynomial $$x^3 - 7x + 6$$ using the Rational Root Theorem. 2. **Recall the Rational Root Theorem:** Possible rational roots are of the form $$\pm \frac{p}{q}$$ where $$p$$ divides the constant term and $$q$$ divides the leading coefficient. 3. For $$x^3 - 7x + 6$$, the constant term is 6 and the leading coefficient is 1. 4. Factors of 6 are $$\pm1, \pm2, \pm3, \pm6$$ and factors of 1 are $$\pm1$$. 5. Possible rational roots are $$\pm1, \pm2, \pm3, \pm6$$. 6. Test each root by substitution: - $$f(1) = 1^3 - 7(1) + 6 = 1 - 7 + 6 = 0$$, so $$1$$ is a root. - $$f(-1) = (-1)^3 -7(-1) + 6 = -1 + 7 + 6 = 12$$, not zero. - $$f(2) = 8 - 14 + 6 = 0$$, so $$2$$ is a root. - $$f(-2) = -8 + 14 + 6 = 12$$, not zero. - $$f(3) = 27 - 21 + 6=12$$, not zero. - $$f(-3) = -27 + 21 + 6=0$$, so $$-3$$ is a root. - $$f(6)$$ and $$f(-6)$$ are unnecessary after finding three roots for degree 3 polynomial. 7. Since polynomial degree is 3, and we found three roots $$1, 2, -3$$, these are all the roots. **Final answer:** The possible real roots are $$1, 2, -3$$ and indeed they are roots of the polynomial.