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

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: Find the nth-degree polynomial function with real coefficients satisfying the given conditions. n=4; i and 5i are zeros; f(-1)=52       Log On


   

Question 763112: Find the nth-degree polynomial function with real coefficients satisfying the given conditions.
n=4; i and 5i are zeros; f(-1)=52

Found 2 solutions by MathLover1, tommyt3rd:
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!

You are given three roots:
i and 5i
since i is a root, so must its conjugate -i also be a root,
since 5i is a root, so must its conjugate -5i also be a root
You can find f%28x%29 by using zero product rule
f%28x%29=%28x%2Bi%29%28x-i%29%28x%2B5i%29%28x-5i%29+
f%28x%29=%28x%5E2-i%5E2%29%28x%5E2-%285i%29%5E2%29
f%28x%29=%28x%5E2-%28-1%29%29%28x%5E2-25i%5E2%29
f%28x%29=%28x%5E2%2B1%29%28x%5E2-25%28-1%29%29
f%28x%29=%28x%5E2%2B1%29%28x%5E2%2B25%29
f%28x%29=x%5E4%2B25x%5E2%2Bx%5E2%2B25
f%28x%29=x%5E4%2B26x%5E2%2B25....your polynomial

check if

f%28-1%29=52
f%28-1%29=%28-1%29%5E4%2B26%28-1%29%5E2%2B25
f%28-1%29=1%2B26%2A1%2B25
f%28-1%29=52

Answer by tommyt3rd(5050) About Me  (Show Source):
You can put this solution on YOUR website!
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


:)