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) About Me  (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)
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Cheers,
Stan H.
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Answer by MathTherapy(10555) About Me  (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.
matrix%281%2C3%2C+f%28x%29%2C+%22=%22%2C+a%28x+%2B+3%29%28x%5E2+-+6x+%2B+9+%2B+25%29%29
matrix%281%2C3%2C+f%28x%29%2C+%22=%22%2C+a%28x+%2B+3%29%28x%5E2+-+6x+%2B+34%29%29
a = 1, so we get:

Answer by greenestamps(13203) About Me  (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

a%28x-%28-3%29%29%28x-%283%2B5i%29%29%28x-%283-5i%29%29
a%28x%2B3%29%28%28x-3%29-5i%29%28%28x-3%29%2B5i%29
a%28x%2B3%29%28%28x%5E2-6x%2B9%29%2B25%29
a%28x%2B3%29%28x%5E2-6x%2B34%29

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

a%284%29%2829%29+=+116a+=+116
a+=+1

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

f%28x%29+=+%28x%2B3%29%28x%5E2-6x%2B34%29+=+x%5E3-3x%5E2%2B16x%2B102