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Use the given zero to find all the zeros of the function. f(x)= x3+x2+9x+9 zero= r= 3i
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SInce all the coeffiecents are real, -3i, the complex conjugate is also a root.
Now use the fact that the sum of roots of an expression like ax^3+bx^2+cx+d is (-b/a) Let z be the unknown root. so (3i +(-3i) + z) = -1/1
z=-1.
+-3i and -1 are the roots. It is a cubic equation so we are looking for three roots, we have three, so it's done.
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