1. **State the problem:** Given the polynomial equation $$x^4 + 2x^3 - 10x^2 - 18x + 9 = 0$$, we need to: a. List all possible rational roots using the Rational Root Theorem. b. Use synthetic division to test possible roots and find one actual root. c. Use the found root to solve the equation completely. 2. **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. - Constant term: 9 (factors: 1, 3, 9) - Leading coefficient: 1 (factors: 1) Thus, possible rational roots are $$\pm 1, \pm 3, \pm 9$$. 3. **Synthetic division to test roots:** Try $x=1$: Set up synthetic division with root 1: Coefficients: 1 | 2 | -10 | -18 | 9 Carry down 1: Multiply 1 * 1 = 1, add to 2 = 3 Multiply 3 * 1 = 3, add to -10 = -7 Multiply -7 * 1 = -7, add to -18 = -25 Multiply -25 * 1 = -25, add to 9 = -16 Remainder is -16, not zero, so 1 is not a root. Try $x=-1$: Carry down 1: Multiply -1 * 1 = -1, add to 2 = 1 Multiply 1 * -1 = -1, add to -10 = -11 Multiply -11 * -1 = 11, add to -18 = -7 Multiply -7 * -1 = 7, add to 9 = 16 Remainder 16, not zero. Try $x=3$: Carry down 1: Multiply 3 * 1 = 3, add to 2 = 5 Multiply 3 * 5 = 15, add to -10 = 5 Multiply 3 * 5 = 15, add to -18 = -3 Multiply 3 * -3 = -9, add to 9 = 0 Remainder 0, so $x=3$ is a root. 4. **Use root $x=3$ to factor polynomial:** The quotient polynomial is: $$x^3 + 5x^2 + 5x - 3$$ 5. **Solve the cubic:** Use Rational Root Theorem again for $$x^3 + 5x^2 + 5x - 3 = 0$$. Possible roots: $$\pm 1, \pm 3$$. Try $x=1$: $$1 + 5 + 5 - 3 = 8 \neq 0$$ Try $x=-1$: $$-1 + 5 - 5 - 3 = -4 \neq 0$$ Try $x=3$: $$27 + 45 + 15 - 3 = 84 \neq 0$$ Try $x=-3$: $$-27 + 45 - 15 - 3 = 0$$ So $x=-3$ is a root. 6. **Divide cubic by $(x+3)$:** Coefficients: 1 | 5 | 5 | -3 Carry down 1: Multiply -3 * 1 = -3, add to 5 = 2 Multiply -3 * 2 = -6, add to 5 = -1 Multiply -3 * -1 = 3, add to -3 = 0 Quotient: $$x^2 + 2x - 1$$ 7. **Solve quadratic:** $$x^2 + 2x - 1 = 0$$ Use quadratic formula: $$x = \frac{-2 \pm \sqrt{2^2 - 4(1)(-1)}}{2(1)} = \frac{-2 \pm \sqrt{4 + 4}}{2} = \frac{-2 \pm \sqrt{8}}{2} = \frac{-2 \pm 2\sqrt{2}}{2} = -1 \pm \sqrt{2}$$ 8. **Final solutions:** $$x = 3, x = -3, x = -1 + \sqrt{2}, x = -1 - \sqrt{2}$$