Question
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)
I know the conjucants are -i and -3i
(x-i)(x-3i) (x+i)(x=3i)
Answers
Steve
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"
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"