Friday, July 5, 2013

Find the nth-degree polynomial function with real coefficients satisfying the given conditions. n=4; i and 5i are zeros; f(-1)=52

Since coefficients are real this must mean that the complex roots occur in conjugate pairs. We can write: 
%0D%0Af%28x%29=a%28x-i%29%28x%2Bi%29%28x-5i%29%28x%2B5i%29%0D%0A 

and given that f(-1)=52 we can determine a... 

%0D%0Af%28-1%29=a%28-1-i%29%28-1%2Bi%29%28-1-5i%29%281%2B5i%29=52%0D%0A 
a%28%28-1%29%5E2%2B1%5E2%29%28%28-1%29%5E2%2B5%5E2%29=52 

so that 
a%282%29%2826%29=52 

and a=1 
this leads to 
%0D%0Af%28x%29=%28x%5E2%2B1%29%28x%5E2%2B25%29=x%5E4%2B26x%5E2%2B25%0D%0A 


f%28x%29=x%5E4%2B26x%5E2%2B25%0D%0A 


:)

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