Posted by **Julie** on Monday, June 24, 2013 at 3:01am.

how do I find a 4th degree polynomial that has zeros; i,3i with f(-1)=60

I know the conjucants are -i and -3i

(x-i)(x-3i) (x+i)(x=3i)

- College Algebra -
**Steve**, Monday, June 24, 2013 at 4:55am
So, multiply it out to get

f(x) = a(x^2+1)(x^2+9) = a(x^4+10x^2+9)

Now, f(-1) = a(1+10+9) = 20a, so we need a=3 and thus

f(x) = 3(x^4+10x^2+9)

BTW, that's "conjugates"

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