A plotted curve with 4 inflection points is labeled p left parenthesis x right parenthesis. The plot starts in quadrant 3, rises steeply to a turning point in quadrant 2, before dropping and rising twice in quadrant 1.

Question

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 point)
Responses

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 multiplicity of left parenthesis x minus 3 right parenthesis is even because the graph of p ( x ) touches the x -axis at left parenthesis 3 comma 0 right parenthesis and then turns around.

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 left parenthesis x minus 3 right parenthesis is odd because the graph of p ( x ) touches the x -axis at left parenthesis 3 comma 0 right parenthesis and then turns around.

The multiplicity of (x−3) is even because the associated zero, x=3, is odd.
The multiplicity of left parenthesis x minus 3 right parenthesis is even because the associated zero, x equals 3 , is odd.

The multiplicity of (x−3) is odd because the associated zero, x=3, is odd.

Answer 1

The correct response is:

The multiplicity of (x−3) is even because the graph of p(x) touches the x-axis at (3,0) and then turns around.

When a polynomial has a factor with even multiplicity, the graph touches the x-axis at that root but does not cross it; instead, it turns around. In contrast, if the multiplicity were odd, the graph would cross the x-axis at that point. Since the description of the curve indicates that the graph touches the x-axis at (3, 0) and then turns around, this indicates that the multiplicity of (x - 3) is indeed even.