SOLUTION: Find the nth-degree polynomial function with real coefficients satisfying te given conditions. n=3; -5 and 4 +3i are zeros; f(2)= 91. I don't get it. I have no other way to explain

Algebra ->  Rational-functions -> SOLUTION: Find the nth-degree polynomial function with real coefficients satisfying te given conditions. n=3; -5 and 4 +3i are zeros; f(2)= 91. I don't get it. I have no other way to explain      Log On


   

Question 880683: Find the nth-degree polynomial function with real coefficients satisfying te given conditions. n=3; -5 and 4 +3i are zeros; f(2)= 91. I don't get it. I have no other way to explain how I feel about this problem.
Answer by josgarithmetic(39621) About Me  (Show Source):
You can put this solution on YOUR website!
Complex zeros occur as conjugate pairs so you also have the zero, 4-3i. Having f(2)=91, you then have f%28x%29=k%28x-%28-5%29%29%28x-%284%2B3i%29%29%28x-%284-3i%29%29. Simplify this definition.

k%28x%2B5%29%28x-4-3i%29%28x-4%2B3i%29
k%28x%2B5%29%28%28x-4%29%5E2-%283i%29%5E2%29
k%28x%2B5%29%28%28x-4%29%5E2-%28-1%299%29
k%28x%2B5%29%28%28x-4%29%5E2%2B9%29
k%28k%2B5%29%28x%5E2-8x%2B16%2B9%29
k%28x%2B5%29%28x%5E2-8x%2B25%29

For that last condition, f%28x%29=k%28x%2B5%29%28x%5E2-8x%2B25%29, evaluate at x=2, and solve for k.
k%282%2B5%29%28%282%29%5E2-8%2A2%2B25%29=91
k%287%29%284-64%2B25%29=91
k%2A7%2A%2825-60%29=91
k%2A7%2A%28-35%29=91
k=13%2F%28-35%29
k=-%2813%2F35%29
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Function is highlight%28f%28x%29=-%2813%2F35%29%28x%2B5%29%28x%5E2-8x%2B25%29%29;
put into general form through multiplication if you want.