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Understanding the multiplicity of zeros in a polynomial is crucial because it gives us insight into the behavior of the function at those points. When a polynomial is factored, such as with the polynomial given in the exercise, the exponents of the factors determine the multiplicity of the zeros.
For instance, if a factor is raised to the power of 2, like \(x^2\), the corresponding zero \(x = 0\) is said to have a multiplicity of 2.
This means that the graph of the polynomial touches or is tangent to the x-axis at this point and turns around without crossing it.
For a factor like \((2x + 3)^5\), the zero \(x = -\frac{3}{2}\) has a multiplicity of 5.
Here, the graph not only touches the axis but does so in a more flattened manner compared to a lower multiplicity. A multiplicity of 1 would indicate the graph crosses the axis.

• Even multiplicity: The curve touches the x-axis and turns back.
• Odd multiplicity: The curve crosses the x-axis.

This understanding helps us predict the shape and turns of the graph at the zeros.

Factoring polynomials is a foundational skill in algebra that allows us to simplify equations, making them easier to solve. A factored polynomial is represented as a product of its factors, which are typically simpler expressions.
In the exercise, the polynomial \(f(x) = x^2 (2x+3)^5 (x-4)^2\) is already given in its factored form.
Each factor corresponds to a zero of the polynomial, which can be found by setting each factor equal to zero.

• First Factor: \(x^2 = 0\) leads to the zero \(x = 0\).
• Second Factor: \((2x+3)^5 = 0\) leads to the zero \(x = -\frac{3}{2}\).
• Third Factor: \((x-4)^2 = 0\) leads to the zero \(x = 4\).

Identifying the factors correctly is the first and most important step in solving a polynomial equation.

Solving polynomial equations involves finding all the values of \(x\) for which the polynomial equals zero. Once a polynomial is factored, solving the equation is straightforward. Remember, each factor gives us a potential solution or zero of the polynomial.
Take the provided exercise: we broke down the factors to get zeros at \(x=0\), \(x=-\frac{3}{2}\), and \(x=4\).
Here's a recap of how each was found:

• For the factor \(x^2 = 0\), solve to find \(x = 0\).
• For the factor \(2x + 3 = 0\), rearrange to \(2x = -3\), then divide by 2 to get \(x = -\frac{3}{2}\).
• For the factor \(x - 4 = 0\), solve to get \(x = 4\).

After solving these individually, consider the multiplicity of each zero, which tells you how many times each solution occurs.
This process of solving is vital because it gives the specific values for which the polynomial is equal to zero, enhancing understanding and facilitating graphing.