Question

N=3, zeros 4 and 2i, f(2)=-48 find nth degree polinomonal function

Answer 1

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 

$\frac{c}{a}=(2i)(-2i)$
= -4(i²=-1)
=5 

$\frac{b}{a}=-(2i+(-2i))$
=0 

Therefore the equation is; a($x^{2}$+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 :)