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

Find the polynomial function
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2 Answers

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
     
        
   
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