1. The problem asks us to analyze the polynomial \(P(x) = (x+2)^2 (x+3)^2 (x-1)^4 (2x+1)\), which is already factored. 2. First, identify the zeros (roots) of the polynomial by setting each factor equal to zero: - \(x+2=0 \Rightarrow x=-2\) - \(x+3=0 \Rightarrow x=-3\) - \(x-1=0 \Rightarrow x=1\) - \(2x+1=0 \Rightarrow x=-\frac{1}{2}\) 3. Determine the multiplicity of each root to understand the behavior of the graph at these points: - Root \(x=-2\) has multiplicity 2 (even), so the graph touches the x-axis and bounces off. - Root \(x=-3\) has multiplicity 2 (even), same behavior as above. - Root \(x=1\) has multiplicity 4 (even), the graph touches and flattens at this root. - Root \(x=-\frac{1}{2}\) has multiplicity 1 (odd), so the graph crosses the x-axis straight through. 4. The degree of the polynomial is \(2 + 2 + 4 + 1 = 9\), which is odd. 5. The leading term can be found by multiplying the leading terms of each factor: - From \((x+2)^2\), leading term is \(x^2\) - From \((x+3)^2\), leading term is \(x^2\) - From \((x-1)^4\), leading term is \(x^4\) - From \((2x+1)\), leading term is \(2x\) Multiplying: \(x^2 \times x^2 \times x^4 \times 2x = 2x^{9}\) 6. Since the leading coefficient \(2 > 0\) and degree 9 is odd, the graph falls to \(-\infty\) as \(x \to -\infty\) and rises to \(+\infty\) as \(x \to +\infty\). 7. Summary: - Roots at \(-3, -2, -\frac{1}{2}, 1\) with corresponding multiplicities. - Graph touches x-axis at \(-3, -2, 1\) and crosses at \(-\frac{1}{2}\). - End behavior determined by leading term \(2x^9\). This information can be used to sketch the graph accurately.