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)
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Answer:
of iv
+
2
(
4
−
)
=
2
+
6
−
3
x+2({\color{#c92786}{4-x}})=2x+6-3x
x+2(4−x)=2x+6−3x
+
2
(
−
+
4
)
=
2
+
6
−
3
Step-by-step explanation:
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Answer:
(1,0), (0,0),(-2<0)<(-1, -2)
Step-by-step explanation:
You'd have to add 19w in order to cancel out the -19w