Real Zeros 98Dc5D

1. **State the problem:** Find all real zeros of the polynomial $$f(x) = 3x^3 + 14x^2 + 7x - 4$$ given that one zero is $x = -4$. 2. **Recall the Factor Theorem:** If $x = a$ is a zero of $f(x)$, then $(x - a)$ is a factor of $f(x)$. 3. Since $x = -4$ is a zero, $(x + 4)$ is a factor. Use polynomial division or synthetic division to divide $f(x)$ by $(x + 4)$. 4. **Synthetic division setup:** - Coefficients: $3$, $14$, $7$, $-4$ - Divide by root $-4$ Perform synthetic division: $$egin{array}{r|rrrr} -4 & 3 & 14 & 7 & -4 \\ & & -12 & -8 & 4 \\ \\hline & 3 & 2 & -1 & 0 \\ \end{array}$$ Remainder is $0$, confirming $(x + 4)$ is a factor. 5. The quotient polynomial is $$3x^2 + 2x - 1$$. 6. **Find zeros of the quadratic:** Solve $$3x^2 + 2x - 1 = 0$$ using the quadratic formula: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 3 \cdot (-1)}}{2 \cdot 3} = \frac{-2 \pm \sqrt{4 + 12}}{6} = \frac{-2 \pm \sqrt{16}}{6}$$ 7. Simplify the square root: $$x = \frac{-2 \pm 4}{6}$$ 8. Calculate the two roots: - $$x = \frac{-2 + 4}{6} = \frac{2}{6} = \frac{1}{3}$$ - $$x = \frac{-2 - 4}{6} = \frac{-6}{6} = -1$$ 9. **List all real zeros:** $$\boxed{-4, -1, \frac{1}{3}}$$ These are the real zeros of $f(x)$.