precalculus
posted by ladybug .
form a polynomial f(x) with real coefficients having the given degree and zeros. degree 5; zeros 7; i;9+i
enter the polynomial.
f(x)=a(?)

Complex numbers always appear as conjugate pairs, so if you have i, then you also have +i
and if you have 9+i, there will also be 9  i
so we know we have factors of (x+7) , (x^2 + 1) and two more
I will use the sum and product rule to find the other
sum of 9+i and 9  i = 18
product of the above is 81  i^2 = 81 + 1 = 82
resulting in the quadratic factor
x^2 + 18x + 82
so f(x) = (x+7)(x^2 + 1)(x^2 + 18x + 82)
notice, if expanded this will give you a 5th degree polynomial. If you have to expand it, do it very carefully and patiently.
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