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

degrees 5; zeros 1; -i; 7+i

please show all work. I am having a hard time understanding this.

To form a polynomial with real coefficients given the degree and zeros, we need to use the concept of conjugate pairs.

Given that the degree of the polynomial is 5 and the zeros are 1, -i, and 7+i, we can deduce that conjugates of these complex zeros are also zeros.

The first zero is 1, which means x - 1 is a factor of the polynomial.
Next, we have -i as a zero. Since we want real coefficients, the conjugate of -i, which is i, will also be a zero. Hence, (x + i) and (x - i) are factors of the polynomial.
Finally, we have 7+i as a zero, which means the conjugate, 7-i, will also be a zero. Therefore, (x - (7+i)) and (x - (7-i)) are factors of the polynomial.

To form the polynomial, we multiply all the factors together:

f(x) = (x - 1)(x + i)(x - i)(x - (7+i))(x - (7-i))

Let's simplify this:

(x - 1)(x + i)(x - i)(x - (7+i))(x - (7-i))
= (x - 1)(x^2 + x(i - i) + i^2)(x - (7+i))(x - (7-i))
= (x - 1)(x^2 - xi + i^2)(x - (7+i))(x - (7-i))
= (x - 1)(x^2 + 1)(x - (7+i))(x - (7-i))

Further expansion:

(x - 1)(x^2 + 1)(x - (7+i))(x - (7-i))
= (x - 1)(x^2 + 1)((x - 7) - i)((x - 7) + i)
= (x - 1)(x^2 + 1)((x - 7)^2 - i^2)
= (x - 1)(x^2 + 1)((x - 7)^2 + 1)

Hence, the polynomial f(x) with real coefficients and the given zeros and degree is:
f(x) = (x - 1)(x^2 + 1)((x - 7)^2 + 1)