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use the given zero to find all zeros of the function h(x)=3x^3-4x^2+8x+8 zero: 1-√(3i)
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one root will be the compliment of the complex number that is 1+√(3i)
another root is probably real a factor of 8 divided by 3 could be a root h(2/3) = 8/9 - 4*4/9 + 16/3 + 8 = not = 0
use this sum of roots of ax^3 + bx^2 +cx + d = -b/a
try -2/3 i'll use my calculator - yes thats the one -2/3 is a zero so all zeros are 1+√(3i), 1-√(3i) and -2/3
you can often make a good guess of the zero by considering the coefficient of x^3 and last number - here they are 3 and 8
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