Answer:
[tex]P(x)=-\frac{x^3}{2}-x^2-\frac{x}{2}+2[/tex]
Step-by-step explanation:
A polynomial can be written as:
[tex]P(x)=a*(x-x_1)^t*(x-x_2)^v*...*(x-x_n)^w[/tex]
Where [tex]a[/tex] is the lead coefficient. And [tex]x_1,x_2,...,x_n[/tex] are the roots of the polynomial.
And [tex]t,u,w[/tex] are the orders of multiplicity.
The polynomial has degree 3 and it has 2 factors, this is:
[tex]P(x)=a*(x-x_1)^t*(x-x_2)^v[/tex]
We have a root of multiplicity 2 at x = 1, then:
[tex](x-x_1)^t=(x-1)^2[/tex]
And we have a root of multiplicity 1 at x = − 4:
[tex](x-x_2)^v=(x-(-4))^1=(x+4)[/tex]
Then, [tex]P(x)=a*(x-1)^2*(x+4)[/tex]
And the problem says that the y-intercept is y=-2 this means that when x=0, y=-2
Observation: P(x)=y
Now we have to replace P(x) with x=0:
[tex]P(x)=a*(x-1)^2*(x+4)\\P(0)=a*(0-1)^2*(0+4)\\\\-2=4a\\a=-\frac{1}{2}[/tex]
Then [tex]P(x)=-\frac{1}{2} *(x-1)^2*(x+4)[/tex]
[tex]P(x)=-\frac{1}{2}(( x^2-2x+1)*(x+4))\\\\P(x)=-\frac{1}{2}(x^3+4x^2-2x^2+x+4)\\\\P(x)=-\frac{1}{2}(x^3+2x^2+x+4)\\\\P(x)=-\frac{x^3}{2}-x^2-\frac{x}{2}+2[/tex]
You can see that the polynomial is of degree 3.
The result is [tex]P(x)=-\frac{x^3}{2}-x^2-\frac{x}{2}+2[/tex]
