Question 919829: Write a polynomial function of least degree that has the given zeros
-5, 3+2i and also -1, 5, 2
Found 2 solutions by Edwin McCravy, MathLover1:
Answer by Edwin McCravy(20064)   (Show Source):
You can put this solution on YOUR website! Write a polynomial function of least degree that has the given zeros
-5, 3+2i...
Since 3+2i is a solution, then 3-2i must also be a solution if the resulting
polynomial is to have all real coefficients.
Start with these three equations for "x = "
x = -5, x = 3+2i, and we must also include x = 3-2i
Get 0 on the right of each
x+5 = 0, x-3-2i = 0, x-3+2i = 0
Multiply equals by equals:
(x+5)(x-3-2i)(x-3+2i) = 0
Group the (x-3)'s
(x+5)[(x-3)-2i][(x-3)+2i] = 0
Multiply the last two factors:
(x+5)[(x-3)²-4i²] = 0
(x+5)[(x-3)²-4(-1)] = 0
(x+5)[(x-3)²+4] = 0
(x+5)[(x-3)(x-3)+4] = 0
(x+5)[x²-6x+9+4] = 0
(x+5)[x²-6x+13] = 0
x³-6x²+13x+5x²-30x+65
x³-x²-17x+65 = 0
So the function of least degree with real coefficients
that has those zeros is
f(x) = x³-x²-17x+65
[Note: If the problem did not state "with real coefficients", then
this: f(x) = x²+(2-2i)x-(10i+15), with imaginary coefficients would
be the polynomial with least degree, but I'm almost sure the problem
stated "with real coefficients".]
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...and also -1, 5, 2
Start with these three equations for "x = "
x = -1, x = 5, x = 2
Do the same as above and see if you can get
f(x) = x³-6x²+3x+10
Edwin
Answer by MathLover1(20850)  (Show Source):
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