Form a polynomial f(x) with the real coefficients having the given degree and zeros.

Degree 5; Zeros: -3; -i; -6+i

f(x)=a( )

If the zeroes are a1,a2,a3, etc. then the polynomial is:

p(x) = A (x-a1)(x-a2)(x-a3)....

If all the coefficients are real, then the zeroes come in pairs of complex conjugates. So, in this problem it is given that -i is a zero, and then it follows that i is also a zero.

So then a=0?

What would be the polynomial?

To form a polynomial with the given degree and zeros, we need to find the factors corresponding to each zero.

First, let's start with the real zero: -3. For a real zero x = -3, the corresponding factor is (x - (-3)) = (x + 3).

Next, we have the complex zero: -i. Complex zeros always come in conjugate pairs, so the other complex zero is i (the conjugate of -i). For a complex zero of the form x = a + bi (where a is the real part and b is the imaginary part), the corresponding factor is (x - (a + bi)) = (x - a - bi).

For the complex zero -6 + i, the other complex conjugate zero is -6 - i. So we have two factors: (x - (-6 + i)) = (x + 6 - i) and (x - (-6 - i)) = (x + 6 + i).

Now, we can form the polynomial f(x) by multiplying all the factors together:
f(x) = a(x + 3)(x + 6 - i)(x + 6 + i)

Since the degree of f(x) is 5, we know that there must be one more real or complex zero, as the degree and number of zeros are equal in a polynomial.

The above expression is in factored form, so to find the polynomial, we can multiply it out:

f(x) = a(x + 3)(x + 6 - i)(x + 6 + i)
= a(x + 3)(x^2 + 6x + 6x + 36 - ix + ix - i^2)
= a(x + 3)(x^2 + 12x + 36 + 1)
= a(x + 3)(x^2 + 12x + 37)

Hence, the polynomial f(x) = a(x + 3)(x^2 + 12x + 37), where a is any real number.