SOLUTION: Find an​ nth-degree polynomial function with real coefficients satisfying the given conditions. If you are using a graphing​ utility, use it to graph the function and v

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Question 1068275: Find an​ nth-degree polynomial function with real coefficients satisfying the given conditions. If you are using a graphing​ utility, use it to graph the function and verify the real zeros and the given function value.
n=4
-1,4 and 3+4i are zeros
f(1)=-240
solve f(x)=...........
Found 2 solutions by Fombitz, Boreal:
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
Polynomials with real coefficients with complex roots must have complex conjugate roots so,
f%28x%29=a%28x%2B1%29%28x-4%29%28x-%283%2B4i%29%29%28x-%283-4i%29%29
f%28x%29=a%28x%2B1%29%28x-4%29%28x%5E2-6x%2B25%29
So,
f%281%29=a%281%2B1%29%281-4%29%281-6%2B25%29=-240
-120a=-240
a=2
So,
highlight%28f%28x%29=2%28x%2B1%29%28x-4%29%28x%5E2-6x%2B25%29%29
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Answer by Boreal(15235) About Me  (Show Source):
You can put this solution on YOUR website!
-1,4 and 3+4i, and 3-4i are zeros--complex roots are conjugate
(x+1)(x-4) are factors
for 3+4i to be a root, we need 6+/- sqrt (8i), because it is all divided by 2
Therefore, b=-6, and b^2-4ac=-64
4ac=-100, if a=1, c=25
x^2-6x+25 is the other part.
a(x+1)(x-4)(x^2+6x+25) is the polynomial
when x=1, f(1)=-240
a(2)(-3)(32)=-240
-192a=240
a=5/4
(5/4)(x+1)(x-4)(x^2+6x+25)=(5/4)[x^4+3x^3+3x^2-99x-100]