Form a polynomial f(x) with real coefficents having the given degree and zeros degree 5; zeros: 5; -i; -9 +i

f(x)=a

recall that complex roots come in conjugate pairs, so the complete factorization is

(x-5)(x+i)(x-i)(x-(-9+i))(x-(-9-i))
(x-5)(x^2+1)((x+9)-i)((x+9)-i)
(x-5)(x^2+1)((x+9)^2+1)
(x-5)(x^2+1)(x^2+18x+82)
x^5+13x^4-7x^3-397x^2-8x-410

To form a polynomial with the given zeros, we need to use the fact that complex zeros occur in conjugate pairs. The polynomial will have a degree equal to the number of zeros, which in this case is 5.

Let's start by multiplying the factors corresponding to the real zeros:
f(x) = (x - 5)(x - 5)

Next, multiply the factors corresponding to the complex zeros:
f(x) = (x - 5)(x - 5)(x + i)(x - i)

Now let's multiply the remaining factor corresponding to the complex zero -9 + i:
f(x) = (x - 5)(x - 5)(x + i)(x - i)(x - (-9 + i))

Simplify this expression:
f(x) = (x - 5)(x - 5)(x + i)(x - i)(x + 9 - i)

Finally, let's expand and simplify the expression to get the polynomial in standard form:
f(x) = (x^2 - 10x + 25)(x^2 + 1)(x + 9 - i)
= (x^2 - 10x + 25)(x^3 + 9x^2 + (1 - 9i)x + 9 - i)

So, the polynomial f(x) with the given zeros and real coefficients is:
f(x) = (x^2 - 10x + 25)(x^3 + 9x^2 + (1 - 9i)x + 9 - i)