SOLUTION: nth degree polynomial function w/real coefficients satisfying the given conditions n=3 -3 and 1+5i are 0 f(2)=130 f(x)=

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Question 972920: nth degree polynomial function w/real coefficients satisfying the given conditions
n=3
-3 and 1+5i are 0
f(2)=130
f(x)=
Found 2 solutions by josgarithmetic, KMST:
Answer by josgarithmetic(39616) About Me  (Show Source):
You can put this solution on YOUR website!
-3 and 1+5i ARE ZEROS.
This must also include 1-5i.

Start with f%28x%29=k%28x%2B3%29%28x-%281%2B5i%29%29%28x-%281-5i%29%29, and simplify into general form, but keep the unknown factor k, UNDISTRIBUTED.

Next step from the raw factored form,
k%28x%2B3%29%28%28x-1%29-5i%29%28%28x-1%29%2B5i%29
and then
k%28x%2B3%29%28%28x-1%29%5E2-%285i%29%5E2%29
Keep going...

Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
A polynomial function of degree n=3 has 3 complex zeros.
If a polynomial function w/real coefficients has a non-real complex zero,
then the conjugate complex number is also a zero.
Then, the complex zeros of f%28x%29 are -3 , 1%2B5i , and 1-5i .
So the factored form of f%28x%29 is
f%28x%29=a%28x-%28-3%29%29%28x-%281%2B5i%29%29%28x-%281-5i%29%29
f%28x%29=a%28x%2B3%29%28x-1-5i%29%28x-1%2B5i%29%29
f%28x%29=a%28x%2B3%29%28%28x-1%29%5E2-%285i%29%5E2%29%29
f%28x%29=a%28x%2B3%29%28x%5E2-2x%2B1-25%28i%29%5E2%29%29
f%28x%29=a%28x%2B3%29%28x%5E2-2x%2B1-25%28-1%29%29
f%28x%29=a%28x%2B3%29%28x%5E2-2x%2B1%2B25%29
f%28x%29=a%28x%2B3%29%28x%5E2-2x%2B26%29
So, 130=f%282%29=a%282%2B3%29%282%5E2-2%2A2%2B26%29-->130=5a%284-4%2B26%29-->130=5a%2826%29-->130=130a-->a=1 .
So, f%28x%29=%28x%2B3%29%28x%5E2-2x%2B26%29 in factored form,
and if that's not the required form, we can multiply and get