Find the equation of a polynomial function of lowest degree with real coefficients that has 6 and 3 - 2i as solutions.

Question

owlfred · Answer

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amistre64 · Answer

(x-6)(x-(3-2i))

amistre64 · Answer

(x-6)(x-3+2i)

x^2 -3x +2xi
      -6x           -12i+18
--------------------
x^2 -9x +18 +(2xi -12i)

anonymous · Answer

amistre: I dont think that this is with real coefficients

amistre64 · Answer

im working on it lol

anonymous · Answer

I got it now i just forgot the first step

anonymous · Answer

ok, cool :-)

amistre64 · Answer

cool; was I close?

amistre64 · Answer

i think we square this part tho right?  3 - 2i

amistre64 · Answer

its a cubic; with a bend above the line..

amistre64 · Answer

http://www.wolframalpha.com/input/?i=%28x-6%29%28x-3%2B2i%29^2 might be useful

anonymous · Answer

Well when you have (3-2i) my definition you also have the conjugate (3+2i)

anonymous · Answer

*by definition*

anonymous · Answer

if your curious it was x^3-12x^2+49x-78

anonymous · Answer

f(x) = "  "

amistre64 · Answer

http://www.wolframalpha.com/input/?i=%28x-6%29%28x-%283-2i%29%29%28x-%283%2B2i%29%29

amistre64 · Answer

i was curious :)  but yeah, thats what I came up with eventually