Polynomial Construction 4C5B3C

1. **State the problem:** We need to construct a degree 6 polynomial function $f(x)$ with zeros at $x=-1$ (multiplicity 3), $x=2$ (multiplicity 2), and $x=-4$ (multiplicity 1). The graph passes through $(1,-10)$ and as $x \to \infty$, $f(x) \to -\infty$. 2. **Write the general form using zeros and multiplicities:** $$f(x) = a(x+1)^3 (x-2)^2 (x+4)$$ where $a$ is a constant to be determined. 3. **Determine the leading term and sign of $a$:** - The degree is $3 + 2 + 1 = 6$, which matches the requirement. - The leading term comes from multiplying the highest powers: $$a \cdot x^3 \cdot x^2 \cdot x = a x^6$$ - Since as $x \to \infty$, $f(x) \to -\infty$, the leading coefficient $a$ must be negative. 4. **Use the point $(1,-10)$ to find $a$:** Substitute $x=1$, $f(1) = -10$: $$-10 = a(1+1)^3 (1-2)^2 (1+4)$$ Calculate each factor: $$(1+1)^3 = 2^3 = 8$$ $$(1-2)^2 = (-1)^2 = 1$$ $$(1+4) = 5$$ So: $$-10 = a \times 8 \times 1 \times 5 = 40a$$ Solve for $a$: $$a = \frac{-10}{40} = -\frac{1}{4}$$ 5. **Write the specific polynomial:** $$f(x) = -\frac{1}{4} (x+1)^3 (x-2)^2 (x+4)$$ 6. **Verify conditions:** - Degree: $3 + 2 + 1 = 6$ correct. - Zeros and multiplicities: - At $x=-1$, multiplicity 3. - At $x=2$, multiplicity 2. - At $x=-4$, multiplicity 1. - Passes through $(1,-10)$ by construction. - Leading coefficient negative, so as $x \to \infty$, $f(x) \to -\infty$. All conditions are satisfied. **Final answer:** $$f(x) = -\frac{1}{4} (x+1)^3 (x-2)^2 (x+4)$$