Find the zeros of the given polynomial function and give the multiplicity of each. State whether the graph crosses or touches the x-axis at each zero. $f\left(x\right)=x^2\left(x-1\right)^3\left(2x+6\right)$

• A. Cross at $x=0,$ Cross at $x=1,$ Cross at $x=3$
• B. Touch at $x=0,$ Cross at $x=-1,$ Cross at $x=3$
• C. Cross at $x=0,$ Touch at $x=1,$ Touch at $x=-3$
• D. Touch at $x=0,$ Cross at $x=1,$ Cross at $x=-3$

Answer: (D)

Verified step by step guidance

1. Identify the factors of the polynomial function f(x) = x^2(x-1)^3(2x+6). The factors are x^2, (x-1)^3, and (2x+6).
2. Set each factor equal to zero to find the zeros of the polynomial: x^2 = 0, (x-1)^3 = 0, and 2x+6 = 0.
3. Solve each equation for x: For x^2 = 0, x = 0. For (x-1)^3 = 0, x = 1. For 2x+6 = 0, x = -3.
4. Determine the multiplicity of each zero: x = 0 has multiplicity 2, x = 1 has multiplicity 3, and x = -3 has multiplicity 1.
5. Analyze the behavior of the graph at each zero: If the multiplicity is odd, the graph crosses the x-axis. If the multiplicity is even, the graph touches the x-axis. Therefore, the graph touches at x = 0, crosses at x = 1, and crosses at x = -3.