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

Degree 4; zeros:-3 +5i; 2 multiplicity 2

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

Well, let's see if I can entertain you with a polynomial that fits the criteria. We know that complex zeros come in conjugate pairs, so if -3 + 5i is a zero, then its conjugate -3 - 5i must also be a zero.

Now, let's consider the multiplicity of 2 at 2. If a zero has a multiplicity of 2, it means it appears twice. So, we can write (x - 2)(x - 2) to represent the two occurrences of the zero 2.

Putting it all together, we have:

f(x) = a(x - (-3 + 5i))(x - (-3 - 5i))(x - 2)(x - 2).

To simplify it a bit, we can further expand it:

f(x) = a(x + 3 - 5i)(x + 3 + 5i)(x - 2)(x - 2).

An unorthodox polynomial indeed, but trust me, it'll function just fine!

To form the polynomial with the given degree and zeros, we need to consider the complex zeros as conjugate pairs. Let's break it down step by step.

1. The degree of the polynomial is 4, so the polynomial will have 4 roots or zeros.

2. The given zeros are -3 + 5i and 2 (with a multiplicity of 2).

3. Since -3 + 5i is a complex number, its conjugate is also a root. So, the conjugate of -3 + 5i is -3 - 5i. Therefore, the complex zeros are -3 + 5i, -3 - 5i, and 2 (with a multiplicity of 2).

4. Since the polynomial has real coefficients, complex zeros always occur in conjugate pairs. This means that if a + bi is a zero, then a - bi is also a zero.

5. Now, we can write the polynomial using the zero product property. The general form of the polynomial with given zeros is:
f(x) = a(x - z1)(x - z2)(x - z3)(x - z4),
where z1, z2, z3, and z4 are the zeros.

Substituting our zeros, we have:
f(x) = a(x - (-3 + 5i))(x - (-3 - 5i))(x - 2)(x - 2)

Simplifying further:
f(x) = a(x + 3 - 5i)(x + 3 + 5i)(x - 2)(x - 2)

Now, we can expand the equation to get the final polynomial form. Remember, (a + b)(a - b) = a^2 - b^2:
f(x) = a[(x + 3)^2 - (5i)^2][(x - 2)^2]

Simplifying the square terms:
f(x) = a[(x^2 + 6x + 9) - 25][(x - 2)^2]

Combining like terms:
f(x) = a[(x^2 + 6x - 16)(x - 2)^2]

This is the polynomial f(x) with real coefficients and the given zeros.

degree 4

zeros 3-5i
-5 multiplicity 2
enter the polynomial f(x)=a()

since complex roots come in pairs, you have

(x-(-3+5i))(x-(-3-5i))(x-2)^2
((x+3)+5i)((x+3)-5i)(x-2)^2
(x^2 + 6x + 34)(x^2 - 4x + 4)
x^4 + 2x^3 + 14x^2 - 112x + 136