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July 29, 2014

July 29, 2014

Posted by **Jack** on Wednesday, July 15, 2009 at 4:47pm.

f(x) = x^3 + x^2 + 9x + 9 r = 3i

I do not understand this!!

- Math -
**Reiny**, Wednesday, July 15, 2009 at 7:13pmthe 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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