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 1129188: 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=3;
-3 and 3+5i are zeros;
f (1) = 116
f(x)=
Found 3 solutions by stanbon, MathTherapy, greenestamps:
Answer by stanbon(75887)   (Show Source): You can put this solution on YOUR website!
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=3;
-3 and 3+5i are zeros;
f (1) = 116
----
Since the coefficients are Real Numbers 3-5i is also a zero.
----
f(x) = a(x+3)(x-3-5i)(x-3+5i)
f(x) = a(x+3)[(x-3)^2-(25i^2)]
f(x) = a(x+3)(x^2-6x+9+625)
----
Solve for "a"::
116 = a(1+3)(1-6+634)
116 = a(4)(629)
a = 0.046
----
f(x) = 0.46(x+3)(x^2-6x+634)
-------
Cheers,
Stan H.
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Answer by MathTherapy(10555)   (Show Source): You can put this solution on YOUR website!

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=3;
-3 and 3+5i are zeros;
f (1) = 116
f(x)=
IGNORE STANBON'S answer.





a = 1, so we get:

 

Answer by greenestamps(13203)   (Show Source): You can put this solution on YOUR website!
 


The non-real zeros come in conjugate pairs; so the 3 roots are -3, 3+5i,and 3-5i.  Then the polynomial is











To determine the value of the leading coefficient a, use the given fact that f(1)=116:







So the leading coefficient is 1, and the polynomial is found by performing the indicated multiplication.


 

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