1. **State the problem:** Find the roots of the cubic polynomial $$x^3 + 4x^2 - 17x - 60$$ and verify them. 2. **Formula and rules:** To find roots of a cubic polynomial, we can try factoring by grouping or use the Rational Root Theorem to test possible roots. 3. **Apply Rational Root Theorem:** Possible rational roots are factors of the constant term 60 divided by factors of the leading coefficient 1, i.e., $$\pm1, \pm2, \pm3, \pm4, \pm5, \pm6, \pm10, \pm12, \pm15, \pm20, \pm30, \pm60$$. 4. **Test roots:** Substitute $x = -6$: $$(-6)^3 + 4(-6)^2 - 17(-6) - 60 = -216 + 144 + 102 - 60 = -30 \neq 0$$ Try $x = -2$: $$(-2)^3 + 4(-2)^2 - 17(-2) - 60 = -8 + 16 + 34 - 60 = -18 \neq 0$$ Try $x = 5$: $$5^3 + 4(5)^2 - 17(5) - 60 = 125 + 100 - 85 - 60 = 80 \neq 0$$ 5. **Re-examine the problem:** The graph shows roots at $x = -6, -2, 5$. Let's verify carefully: At $x = -6$: $$(-6)^3 + 4(-6)^2 - 17(-6) - 60 = -216 + 144 + 102 - 60 = -30$$ (not zero, so check calculation) Recalculate: $$-216 + 144 = -72$$ $$-72 + 102 = 30$$ $$30 - 60 = -30$$ So $x = -6$ is not a root. At $x = -2$: $$-8 + 16 + 34 - 60 = -18$$ not zero. At $x = 5$: $$125 + 100 - 85 - 60 = 80$$ not zero. 6. **Try synthetic division with $x = 3$:** Coefficients: 1, 4, -17, -60 Bring down 1: Multiply 1*3=3, add to 4=7 Multiply 7*3=21, add to -17=4 Multiply 4*3=12, add to -60=-48 Remainder is -48, not zero. 7. **Try $x = -5$:** Bring down 1: Multiply 1*(-5)=-5, add to 4=-1 Multiply -1*(-5)=5, add to -17=-12 Multiply -12*(-5)=60, add to -60=0 Remainder zero, so $x = -5$ is a root. 8. **Divide polynomial by $(x + 5)$:** $$\frac{x^3 + 4x^2 - 17x - 60}{x + 5} = x^2 - x - 12$$ 9. **Factor quadratic:** $$x^2 - x - 12 = (x - 4)(x + 3)$$ 10. **Roots:** $$x = -5, 4, -3$$ **Final answer:** The roots of the polynomial are $$\boxed{-5, 4, -3}$$.