Zeros Polynomial 78Aa68

1. **State the problem:** Find the zeros of the cubic polynomial $$8x^3 - 42x^2 + 9x + 5$$ and determine their multiplicities. 2. **Recall the formula and rules:** To find zeros of a polynomial, we solve $$8x^3 - 42x^2 + 9x + 5 = 0$$. Zeros can be found by factoring or using the Rational Root Theorem to test possible roots. 3. **Apply Rational Root Theorem:** Possible rational roots are factors of the constant term 5 divided by factors of leading coefficient 8, i.e., $$\pm 1, \pm \frac{1}{2}, \pm \frac{1}{4}, \pm \frac{1}{8}, \pm 5, \pm \frac{5}{2}, \pm \frac{5}{4}, \pm \frac{5}{8}$$. 4. **Test roots:** Substitute $x=1$: $$8(1)^3 - 42(1)^2 + 9(1) + 5 = 8 - 42 + 9 + 5 = -20 \neq 0$$ Substitute $x=\frac{1}{2}$: $$8\left(\frac{1}{2}\right)^3 - 42\left(\frac{1}{2}\right)^2 + 9\left(\frac{1}{2}\right) + 5 = 8\cdot \frac{1}{8} - 42 \cdot \frac{1}{4} + \frac{9}{2} + 5 = 1 - 10.5 + 4.5 + 5 = 0$$ So, $x=\frac{1}{2}$ is a root. 5. **Divide polynomial by $(x - \frac{1}{2})$:** Use synthetic division or polynomial division: Divide $$8x^3 - 42x^2 + 9x + 5$$ by $$x - \frac{1}{2}$$: Multiply divisor by 8 to clear fraction: divide by $$2x - 1$$ instead. Divide $$8x^3 - 42x^2 + 9x + 5$$ by $$2x - 1$$: - Leading term: $$\frac{8x^3}{2x} = 4x^2$$ - Multiply divisor by $$4x^2$$: $$8x^3 - 4x^2$$ - Subtract: $$(8x^3 - 42x^2) - (8x^3 - 4x^2) = -38x^2$$ - Bring down $$+9x$$ - Next term: $$\frac{-38x^2}{2x} = -19x$$ - Multiply divisor by $$-19x$$: $$-38x^2 + 19x$$ - Subtract: $$( -38x^2 + 9x ) - ( -38x^2 + 19x ) = -10x$$ - Bring down $$+5$$ - Next term: $$\frac{-10x}{2x} = -5$$ - Multiply divisor by $$-5$$: $$-10x + 5$$ - Subtract: $$( -10x + 5 ) - ( -10x + 5 ) = 0$$ Quotient is $$4x^2 - 19x - 5$$. 6. **Factor quadratic:** $$4x^2 - 19x - 5 = 0$$ Use quadratic formula: $$x = \frac{19 \pm \sqrt{(-19)^2 - 4 \cdot 4 \cdot (-5)}}{2 \cdot 4} = \frac{19 \pm \sqrt{361 + 80}}{8} = \frac{19 \pm \sqrt{441}}{8} = \frac{19 \pm 21}{8}$$ So, $$x = \frac{19 + 21}{8} = \frac{40}{8} = 5$$ $$x = \frac{19 - 21}{8} = \frac{-2}{8} = -\frac{1}{4}$$ 7. **List all zeros and multiplicities:** - $x = \frac{1}{2}$ (multiplicity 1) - $x = 5$ (multiplicity 1) - $x = -\frac{1}{4}$ (multiplicity 1) All roots are simple (multiplicity 1). **Final answer:** The zeros of $$8x^3 - 42x^2 + 9x + 5$$ are $$x = \frac{1}{2}, 5, -\frac{1}{4}$$ each with multiplicity 1.