Answer:
Factored form: [tex]P(x)=-\frac{1}{1500}(x+20)(x+5)(x-15)[/tex].
Standard form: [tex]P(x)=-\frac{x^3}{1500}-\frac{x^2}{150}+\frac{11x}{60}+1[/tex]
Step-by-step explanation:
The factor form of a polynomial is
[tex]P(x)=a(x-c_1)(x-c_2)...(x-c_n)[/tex]
where, a is a constant and leading coefficient, [tex]c_1,c_2, ...,c_n[/tex] are n zeroes of the polynomial.
From the given graph it is clear that the x-intercepts of the function are -20, -5 and 15. It means the zeroes of the given function are -20, -5 and 15. So, the required function is
[tex]P(x)=a(x-(-20))(x-(-5))(x-15)[/tex]
[tex]P(x)=a(x+20)(x+5)(x-15)[/tex] .... (1)
From the given graph it is clear that y-intercept of the function is (0,1). Use the y-intercept of the graph to find the value of a.
[tex]1=a(0+20)(0+5)(0-15)[/tex]
[tex]1=-1500a[/tex]
Divide both sides by -1500.
[tex]\frac{1}{-1500}=a[/tex]
[tex]-\frac{1}{1500}=a[/tex]
Put [tex]a=-\frac{1}{1500}[/tex] in equation (1).
[tex]P(x)=-\frac{1}{1500}(x+20)(x+5)(x-15)[/tex]
Therefore the factored form of the function is [tex]P(x)=-\frac{1}{1500}(x+20)(x+5)(x-15)[/tex].
Expand the above function to find the standard form of the function
[tex]P(x) = \frac{-x^3 - 10 x^2 + 275 x + 1500}{1500}[/tex]
[tex]P(x)=-\frac{x^3}{1500}-\frac{x^2}{150}+\frac{11x}{60}+1[/tex]
Therefore the standard form of the function is [tex]P(x)=-\frac{x^3}{1500}-\frac{x^2}{150}+\frac{11x}{60}+1[/tex].
