SOLUTION: Form a polynomial f(x) with real coefficients having the given degree and zeros. Degree 4; zeroes: -3-2i; -2 multiplicity 2 So what i've done so far (x+(3-2i))(x+(3+2i))(x

Question 596416: Form a polynomial f(x) with real coefficients having the given degree and zeros.
Degree 4; zeroes: -3-2i; -2 multiplicity 2
So what i've done so far
(x+(3-2i))(x+(3+2i))(x+2)^2
Now my algebra teacher taught me a shortcut on the i's.
(x+3)(-2i)(x+3)(2i)(x+2)^2
(x+3)^2 = x^2 + 6x + 9 / (-2i)(2i) = -4i^2 = 4 / (x+2)^2 = x^2 + 4x + 4
after all of that I have a huge problem just multiplying the quadratics together so I will give you what I had as an answer and if you can guide me in the wrong direction on my sign mistakes.
(x^2+6x+9)(x^2+4x+4)+4
(x^4+4x^3+4x^2+6x^3+24x^2+24x+9x^2+36x+36)
x^4+10x^3+37x^2+60x+36
I have a feeling it's my final multiplication or the negative and positive im getting wrong.

By the way im suppose to get a polynomial answer.
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
Form a polynomial f(x) with real coefficients having the given degree and zeros.
Degree 4 zeroes: -3-2i; -2 multiplicity 2
f(x) = (x-(-3-2i))(x-(-3+2i))(x+2)^2
-------
Rearrange:
f(x) = ((x+3)+2i)((x+3)-2i)(x+2)^2
-------
f(x) = [(x+3)^2+4)(x^2+4x+4)
-----
f(x) = (x^2+6x+13)(x^2+4x+4)
---------------------------------------
Cheers,
Stan H.
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