Find a polynomial function with the zeros −1 (multiplicity 2) 0,3 (multiplicity 2) whose graph passes through the point (1,-48)

Question

Find the polynomial function

Answer 1

Zeros =-1,0,3

 

Factors:(x+1)^2 ,x,(x-3)^2

 

f(x)=a(x)(x+1)^2(x-3)^2

 

graph passes through the point (1,-48)=(x,y)

 

-48=a(1)(1+1)^2(1-3)^2

 

-48=a(1)(2)^2(-2)^2

 

-48=a(1)(4)(4)

 

-48=16a

 

-3=a

 

 

f(x)=a(x)(x+1)^2(x-3)^2

 

f(x)=-3(x)(x+1)^2(x-3)^2

     

        

   

Answer 2

Multiplicity means the number of times a root appears I the function.

 

f(x) = Cx²(x + 1)²(x - 3)²

 

 

Then to find the value of C, set f(x)=-48 at x=1.

 

-48 = C(1)²(1 + 1)²(1 - 3)²

-48 = (4)(4)C

-48 = 16C

-3 = C