From the graph, the real roots are [tex]x=-2 \text{ (twice)}[/tex], and [tex]x=+4[/tex]
The corresponding factors are [tex](x+2)^2 \text{, and }(x-4)[/tex]
The root that has an odd multiplicity is [tex]x=+4[/tex]
The graph touches the x-axis at the point [tex]x=-2[/tex], and crosses it at [tex]x=+4[/tex]. These are the locations of the roots of the polynomial.
The graph is assumed to be of degree three, so the sum of all the multiplicities should be 3. Looking at the roots again,
- The graph crosses the x-axis at the root [tex]x=+4[/tex]. Since the graph looks almost linear at that point, this root must have multiplicity of 1. Thus, this root has an odd multiplicity.
- The graph touches the x-axis at the root [tex]x=-2[/tex]. Since the total multiplicity is three, this root certainly has a multiplicity of 2. So, the multiplicity of this root is even.
So, from the information in the graph we can conclude that the polynomial has the factored form [tex](x+2)^2(x-4)[/tex]. Therefore,
- The factors are [tex](x+2)[/tex] repeated twice, and [tex](x-4)[/tex]
- The roots are [tex]x=-2 \text{ (twice)}[/tex], and [tex]x=+4[/tex].
Learn more about multiplicities of roots here: https://brainly.com/question/11754233
