Posted by Jack on Wednesday, July 15, 2009 at 4:47pm.
the zeros (roots) could be found without the given hint that one of the roots is 3i
f(x) = x^3 + x^2 + 9x + 9
= x^2(x+1) + 9(x+1)
= (x^2+9)(x+1)
for the zeros,
x+1 = 0 or x = -1
or
x^2 + 9 = 0
x^2 = -9
x = ±3√-1
= ±3i
so the roots are ±3i, -1
Using the given hint that 3i is a root, one property of irrational or complex roots is that they must come in "conjugate pairs" to end up with rational coefficients.
so 2 factors would be (x-3i) and (x+3i)
then (x+3i)(x-3i) = x^2 + 9
Using long division, divide your original function by x^2 + 9 to get the other factor of x+1
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