1. **State the problem:** We are given the polynomial $$U(x) = 6x^4 + 35x^3 + 47x^2 - 15x - 9$$ and asked two questions: a) Identify the list of all possible rational roots of $$U(x)$$. b) Factor $$U(x)$$ completely, find all roots, and fill in the blanks in the given factored form and root expressions. 2. **Possible rational roots (Rational Root Theorem):** The possible rational roots of a polynomial with integer coefficients are given by $$\pm \frac{p}{q}$$ where $$p$$ divides the constant term and $$q$$ divides the leading coefficient. - Constant term: $$-9$$, divisors: $$\pm 1, \pm 3, \pm 9$$ - Leading coefficient: $$6$$, divisors: $$\pm 1, \pm 2, \pm 3, \pm 6$$ So possible rational roots are: $$\pm 1, \pm 3, \pm 9, \pm \frac{1}{2}, \pm \frac{3}{2}, \pm \frac{9}{2}, \pm \frac{1}{3}, \pm \frac{3}{3} (= \pm 1), \pm \frac{9}{3} (= \pm 3), \pm \frac{1}{6}, \pm \frac{3}{6} (= \pm \frac{1}{2}), \pm \frac{9}{6} (= \pm \frac{3}{2})$$ Simplifying duplicates, the full list is: $$\pm 1, \pm 3, \pm 9, \pm \frac{1}{2}, \pm \frac{3}{2}, \pm \frac{9}{2}, \pm \frac{1}{3}, \pm \frac{1}{6}$$ 3. **Answer to part (a):** The correct list is: $$\pm 1, \pm 3, \pm 9, \pm \frac{1}{2}, \pm \frac{3}{2}, \pm \frac{9}{2}, \pm \frac{1}{3}, \pm \frac{1}{6}$$ 4. **Factor the polynomial:** We try to find roots by testing possible rational roots. Test $$x = -1$$: $$U(-1) = 6(-1)^4 + 35(-1)^3 + 47(-1)^2 - 15(-1) - 9 = 6 - 35 + 47 + 15 - 9 = 24 \neq 0$$ Test $$x = 1$$: $$U(1) = 6 + 35 + 47 - 15 - 9 = 64 \neq 0$$ Test $$x = -3$$: $$U(-3) = 6(81) + 35(-27) + 47(9) - 15(-3) - 9 = 486 - 945 + 423 + 45 - 9 = 0$$ So $$x = -3$$ is a root. 5. **Divide $$U(x)$$ by $$x + 3$$:** Using synthetic division: Coefficients: 6 | 35 | 47 | -15 | -9 Bring down 6. Multiply by -3: 6 * -3 = -18; add to 35: 17 Multiply 17 * -3 = -51; add to 47: -4 Multiply -4 * -3 = 12; add to -15: -3 Multiply -3 * -3 = 9; add to -9: 0 remainder Quotient: $$6x^3 + 17x^2 - 4x - 3$$ 6. **Factor the cubic $$6x^3 + 17x^2 - 4x - 3$$:** Try rational roots again: possible roots $$\pm 1, \pm \frac{1}{2}, \pm \frac{1}{3}, \pm \frac{1}{6}$$ Test $$x = \frac{1}{3}$$: $$6(\frac{1}{3})^3 + 17(\frac{1}{3})^2 - 4(\frac{1}{3}) - 3 = 6(\frac{1}{27}) + 17(\frac{1}{9}) - \frac{4}{3} - 3 = \frac{6}{27} + \frac{17}{9} - \frac{4}{3} - 3 = \frac{2}{9} + \frac{17}{9} - \frac{12}{9} - \frac{27}{9} = \frac{19 - 12 - 27}{9} = \frac{-20}{9} \neq 0$$ Test $$x = -\frac{1}{3}$$: $$6(-\frac{1}{3})^3 + 17(-\frac{1}{3})^2 - 4(-\frac{1}{3}) - 3 = 6(-\frac{1}{27}) + 17(\frac{1}{9}) + \frac{4}{3} - 3 = -\frac{6}{27} + \frac{17}{9} + \frac{4}{3} - 3 = -\frac{2}{9} + \frac{17}{9} + \frac{12}{9} - \frac{27}{9} = 0$$ So $$x = -\frac{1}{3}$$ is a root. 7. **Divide cubic by $$3x + 1$$:** Synthetic division with root $$x = -\frac{1}{3}$$ corresponds to divisor $$3x + 1$$. Coefficients: 6 | 17 | -4 | -3 Bring down 6. Multiply by -\frac{1}{3}: 6 * -\frac{1}{3} = -2; add to 17: 15 Multiply 15 * -\frac{1}{3} = -5; add to -4: -9 Multiply -9 * -\frac{1}{3} = 3; add to -3: 0 remainder Quotient: $$6x^2 + 15x - 9$$ 8. **Factor quadratic $$6x^2 + 15x - 9$$:** Divide all terms by 3: $$3(2x^2 + 5x - 3)$$ Factor inside parentheses: Find two numbers that multiply to $$2 \times (-3) = -6$$ and add to 5: 6 and -1. Rewrite: $$2x^2 + 6x - x - 3 = 2x(x + 3) - 1(x + 3) = (2x - 1)(x + 3)$$ So full factorization: $$6x^2 + 15x - 9 = 3(2x - 1)(x + 3)$$ 9. **Complete factorization of $$U(x)$$:** $$U(x) = (x + 3)(3)(2x - 1)(x + 3) = 3(x + 3)^2(2x - 1)$$ 10. **Fill in the blanks:** $$U(x) = (x + 3)^2 (3x - 1) (2)$$ but since the factor 3 is outside, rewrite as: $$U(x) = (x + 3)^2 (3x - 1) 2$$ is not standard; better to write as: $$U(x) = 3(x + 3)^2 (2x - 1)$$ The problem states: $$U(x) = (x + \_ )^2 ( \_ x - 1) (A x + \_ )$$ We can write: $$U(x) = (x + 3)^2 (3x - 1) (2)$$ but the factor 2 is a constant, so the form is: $$U(x) = (x + 3)^2 (3x - 1) (2)$$ So the blanks are: - First blank: 3 - Second blank: 3 - A: 3 - Last blank: 1 But the problem's form is ambiguous; assuming the form is: $$(x + 3)^2 (3x - 1)$$ Then: - $$= (x + 3)^2 (3x - 1)$$ - $$A = 3$$ - The roots are: $$x = -3$$ (double root), $$x = \frac{1}{3}$$ and the other root is $$x = 1/\_$$ which is $$x = \frac{1}{3}$$ again, so only three roots total. 11. **Calculate discriminant $$\Delta$$ of quadratic $$3x - 1 = 0$$:** Linear, so no discriminant needed. For the quadratic factor, the discriminant is not needed since it's linear. 12. **Final roots:** $$x = -3$$ (double root), $$x = \frac{1}{3}$$ 13. **Summary:** - Possible rational roots: $$\pm 1, \pm 3, \pm 9, \pm \frac{1}{2}, \pm \frac{3}{2}, \pm \frac{9}{2}, \pm \frac{1}{3}, \pm \frac{1}{6}$$ - Factorization: $$U(x) = 3(x + 3)^2 (2x - 1)$$ - Roots: $$x = -3$$ (double root), $$x = \frac{1}{2}$$ Note: The root from $$2x - 1 = 0$$ is $$x = \frac{1}{2}$$, correcting previous step. So the roots are $$x = -3$$ (multiplicity 2), $$x = \frac{1}{2}$$. The problem's answer format: $$U(x) = (x + 3)^2 (2x - 1) (3)$$ or equivalently $$3(x + 3)^2 (2x - 1)$$ Roots: $$x = -3$$, $$x = \frac{1}{2}$$ The root $$x = 1$$ is not a root. So the blanks: - First blank: 3 - Second blank: 3 - A: 2 - Last blank: 1 - $$\Delta$$ for quadratic $$2x - 1$$ is not applicable (linear). - Roots: $$x = -3$$, $$x = \frac{1}{2}$$