Consider the graph of p(x). Given that (x-3) is a factor of p(x), is the multiplicity of (x-3) even or odd? Why?
1. The multiplicity of (x-3) is odd because the graph of p(x) touches the x-axis at (3,0) and then turns around.
2. The multiplicity of (x-3) is even because the graph of p(x) touches the x-axis at (3,0) and then turns around.
3. The multiplicity of (x-3) is even because the associated zero, x=3, is odd.
4. The multiplicity of (x-3) is odd because the associated zero, x=3, is odd.
2. The multiplicity of (x-3) is even because the graph of p(x) touches the x-axis at (3,0) and then turns around.
The correct answer is:
1. The multiplicity of (x-3) is odd because the graph of p(x) touches the x-axis at (3,0) and then turns around.
The multiplicity of a factor of a polynomial refers to the number of times the factor appears in the polynomial. In this case, if the graph touches the x-axis at (3,0) and then turns around, it means that the factor (x-3) occurs an odd number of times. Therefore, the multiplicity of (x-3) is odd, and option 1 is the correct answer.
To determine the multiplicity of a factor in a polynomial, we need to analyze how the factor affects the graph of the polynomial. In this case, we are given that (x-3) is a factor of p(x), which means that when x=3, p(x) = 0.
To determine whether the multiplicity of (x-3) is even or odd, we need to look at how the graph crosses or touches the x-axis at x=3.
If the graph crosses the x-axis at (3,0) and then turns around, it suggests that the factor (x-3) has an odd multiplicity. This means that the polynomial p(x) includes (x-3) raised to an odd power. Therefore, option 1 is the correct answer: "The multiplicity of (x-3) is odd because the graph of p(x) touches the x-axis at (3,0) and then turns around."