1. **State the problem:** We are given two polynomials: $$f(x) = 2x (x + 3) (x - 2)^2$$ $$g(x) = -x (x - 2)^2 (x + 3)^3$$ We need to sketch sign charts and determine where each polynomial is positive ($>0$) and negative ($<0$). --- 2. **Recall the rules for sign charts:** - The roots (zeros) of the polynomial are where the function equals zero. - The sign of the polynomial changes at roots with odd multiplicity. - The sign does not change at roots with even multiplicity. --- 3. **Analyze $f(x)$:** - Roots: $x=0$ (multiplicity 1), $x=-3$ (multiplicity 1), $x=2$ (multiplicity 2). - Since $(x-2)^2$ has even multiplicity, sign does not change at $x=2$. 4. **Sign intervals for $f(x)$:** - Intervals: $(-\infty, -3)$, $(-3, 0)$, $(0, 2)$, $(2, \infty)$. - Test points: - For $x=-4$ in $(-\infty, -3)$: $2(-4)(-4+3)(-4-2)^2 = 2(-4)(-1)(36) = 288 > 0$. - For $x=-1$ in $(-3, 0)$: $2(-1)(-1+3)(-1-2)^2 = 2(-1)(2)(9) = -36 < 0$. - For $x=1$ in $(0, 2)$: $2(1)(1+3)(1-2)^2 = 2(1)(4)(1) = 8 > 0$. - For $x=3$ in $(2, \infty)$: $2(3)(3+3)(3-2)^2 = 2(3)(6)(1) = 36 > 0$. 5. **Sign chart for $f(x)$:** - Positive on $(-\infty, -3)$, $(0, 2)$, and $(2, \infty)$. - Negative on $(-3, 0)$. 6. **Final solution for $f(x)$:** - $f(x) > 0$ on $(-\infty, -3) \cup (0, \infty)$. - $f(x) < 0$ on $(-3, 0)$. --- 7. **Analyze $g(x)$:** - Roots: $x=0$ (multiplicity 1), $x=2$ (multiplicity 2), $x=-3$ (multiplicity 3). - Sign changes at roots with odd multiplicity: $x=0$ and $x=-3$. - No sign change at $x=2$ (even multiplicity). 8. **Sign intervals for $g(x)$:** - Intervals: $(-\infty, -3)$, $(-3, 0)$, $(0, 2)$, $(2, \infty)$. - Test points: - For $x=-4$ in $(-\infty, -3)$: $-(-4)(-4-2)^2(-4+3)^3 = -(-4)(36)(-1)^3 = -(-4)(36)(-1) = -144 < 0$. - For $x=-1$ in $(-3, 0)$: $-(-1)(-1-2)^2(-1+3)^3 = -(-1)(9)(2)^3 = -(-1)(9)(8) = 72 > 0$. - For $x=1$ in $(0, 2)$: $-(1)(1-2)^2(1+3)^3 = -(1)(1)(4)^3 = -(1)(64) = -64 < 0$. - For $x=3$ in $(2, \infty)$: $-(3)(3-2)^2(3+3)^3 = -(3)(1)(6)^3 = -(3)(216) = -648 < 0$. 9. **Sign chart for $g(x)$:** - Positive on $(-3, 0)$. - Negative on $(-\infty, -3) \cup (0, \infty)$. --- **Summary:** - For $f(x)$: - $f(x) > 0$ on $(-\infty, -3) \cup (0, \infty)$ - $f(x) < 0$ on $(-3, 0)$ - For $g(x)$: - $g(x) > 0$ on $(-3, 0)$ - $g(x) < 0$ on $(-\infty, -3) \cup (0, \infty)$