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

degree 4 zeros 5+3i;3 multiplicity 2

enter the polynomial
f(x)=a?()

see my reply to your other question above

for a zero of 3 with multiplicity of 2
we would have (x-3)(x-3) or (x-3)^2 as part of the function.
For the complex root of 5+3i , there is of course a matching 5-3i root

once you have that , it is simple algebra

To form a polynomial with the given degree and zeros, we need to consider the following points:

1. The degree of the polynomial is 4, which means it will have four terms.

2. The zeros of the polynomial are 5 + 3i (complex zero) and 3 (real zero with multiplicity 2).

To create the polynomial, we need to know that complex zeros always occur in conjugate pairs. It means if 5 + 3i is a zero, then its conjugate 5 - 3i will also be a zero.

To start, we know that for a zero x = a, the polynomial factor will be (x - a). Therefore, for the zero 3 (with multiplicity 2), the factor will be (x - 3)².

To incorporate the complex zeros, we use the conjugate property. So, for (5 + 3i) and (5 - 3i) as zeros, the factors will be (x - (5 + 3i)) and (x - (5 - 3i)) respectively.

Now let's multiply the factors to get the polynomial:

f(x) = (x - 3)² * (x - (5 + 3i)) * (x - (5 - 3i))

Next, let's simplify the equation:

f(x) = (x - 3)² * ((x - 5) - 3i) * ((x - 5) + 3i)

Now we expand and simplify further:

f(x) = (x - 3)² * ((x - 5)² - (3i)²)
= (x - 3)² * ((x - 5)² - 9i²)
= (x - 3)² * ((x - 5)² + 9)

Therefore, the polynomial f(x) with the given degree and zeros is:

f(x) = (x - 3)² * ((x - 5)² + 9)