The degree is 3, the zeros are; 4, 2i, -2i and a point is (-48, 2) 
For zeros; 2i, -2i <-- complex conjugates, always in pairs 

= -4(i²=-1)
=5 

=0 
Therefore the equation is; a(

+5) <-- b value is zero 
Rewrite the equation with all zeros; 
a(x-4)(x²+5)=f(x) <-- put in coordinates of the points to find the value of x
a(2-4)(2²+5)=-48 
a(2)(9)=-48 
a=-48/18 
a=-8/3 
The final polynomial function is; (-8/3)(x-4)(x²+5)=f(x)
Hope I helped :)
 
        
        
        
The answer is: 
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[A]: Erin Naismith lives in Springfield, Massachusetts.
        
             
        
        
        
Answer:
I belive it would be HL because its defenitly not SSS
Step-by-step explanation:
HL Postulate is if the hypotenuse and leg of one right triangle are congruent to the hypotenuse and leg of another right triangle, then the two triangles are congruent. 
 
        
             
        
        
        
Answer:

Step-by-step explanation:
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