1. Problem statement: Sketch the graph and make a sign table for the polynomial $Q(x)=-(x-1)^3(x+2)^2$. 2. Factorization and roots: The factorization is given as $Q(x)=-(x-1)^3(x+2)^2$. 3. Roots and multiplicities: The roots are $x=1$ with multiplicity 3 (odd) and $x=-2$ with multiplicity 2 (even). 4. End behavior: Degree 5 and leading coefficient negative imply $Q(x)\to -\infty$ as $x\to +\infty$ and $Q(x)\to +\infty$ as $x\to -\infty$. 5. Sign analysis: Consider intervals $(-\infty,-2)$, $(-2,1)$, and $(1,\infty)$. 6. Test each interval with a convenient point and compute signs of each factor: For $x=-3$: $(x-1)^3=(-4)^3=-64$ (negative). $(x+2)^2=(-1)^2=1$ (positive). Therefore $Q(-3)= -(-64)(1)=64>0$. 7. For $x=0$: $(x-1)^3=(-1)^3=-1$ (negative). $(x+2)^2=2^2=4$ (positive). Therefore $Q(0)=-(-1)(4)=4>0$. 8. For $x=2$: $(x-1)^3=1^3=1$ (positive). $(x+2)^2=4^2=16$ (positive). Therefore $Q(2)=-(1)(16)=-16<0$. 9. Sign table summary: On $(-\infty,-2)$ positive. At $x=-2$ zero (touches and rebounds). On $(-2,1)$ positive. At $x=1$ zero (crosses with inflection-like behavior). On $(1,\infty)$ negative. 10. Final interpretation and sketch notes: The curve comes from $+\infty$ on the left, touches the x-axis at $x=-2$ and stays above, crosses the x-axis at $x=1$ with inflection-like behavior, and then goes to $-\infty$ as $x\to +\infty$. 11. Final answer: Sign table entries: $Q(x)>0$ for all $x<1$ except at the roots where $Q(-2)=0$ and $Q(1)=0$, and $Q(x)<0$ for $x>1$.