anonymous
  • anonymous
form a third degree polynomial function with real coefficients such that 8+i and 7 are zeros f(x) = ???
Mathematics
schrodinger
  • schrodinger
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anonymous
  • anonymous
Since coefficients are real, the two complex roots have to be conjugates. So the simplest polynomial is \[(x-7)(x -8 -i)(x-8+i)\]
anonymous
  • anonymous
The question wants the answer in a X^3 +bx^2+cX+# ?? so im confused.....
dumbcow
  • dumbcow
x = 8 +- i x-8 = +-i (x-8)^2 = -1 (x-8)^2 +1 = 0 x^2-16x +65 = 0 --> (x-7)(x^2-16x+65) Now distribute

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anonymous
  • anonymous
@dumbcow, why -i if the given is positive(+)?
dumbcow
  • dumbcow
because complex zeros always come in pairs
anonymous
  • anonymous
ahh, yeah. . (+) and (-)
anonymous
  • anonymous
x^3-23x^2-57x-455? is that right?
dumbcow
  • dumbcow
everything except the "-57x" term
anonymous
  • anonymous
177x?
dumbcow
  • dumbcow
yep:)
anonymous
  • anonymous
thanks for the direction much appreciated

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