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 :)
For zeros; 2i, -2i <-- complex conjugates, always in pairs
= -4(i²=-1)
=5
=0
Therefore the equation is; a(
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 :)