calculus
posted by shannon .
p(x)=5x^34x^2+2016
use synthetic division to show that x=2i is a zero of p(x)

Your function, which I will assume was to be
p(x) = 5x^3  4x^2 + 20x  16 , factors very nicely by grouping
p(x) = 5x^3  4x^2 + 20x  16
= x^2(5x4) + 4(5x4)
= (5x4)(x^2 + 4)
so the zeros are
x = 4/5 and x = ± 2i
Using synthetic division for this expression is totally out of place. 
lim x¨‡ 2x 3 +5x 2 −x 3 =
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