(a) How do you form a polynomial $f(x)$ with real coefficients having given degree and zeros? Degree: $4$ ; Zeroes: $ - 5 + 2i$, $3$ multiplicity $2$.
(b) How do you form a polynomial $f(x)$ with real coefficients having a given degree and zeros? Degree: $5$ ; Zeroes: $ - 4, - i, - 3 + i$.

Hint: We are given some zeroes of the polynomials and we are also given the degree of the polynomial. By fundamental theorem of polynomials, we know that there are the same number of zeros in a polynomial, numerically equal to the degree of the polynomial. That is, a polynomial of degree $4$ will have $4$ zeroes, one of degree $5$ will have $5$ zeroes. Also, if a polynomial with real coefficients has a complex number as its zero, then the conjugate of the complex number is also a zero of the polynomial. That is, if $a + ib$ is a zero of a polynomial with real roots, then, $a - ib$ is also a root of the same polynomial. On determining all the zeros, we can find the polynomial, for example, let $\alpha ,\beta ,\nu $ be the zeros of a third degree polynomial, then the polynomial is, $(x - \alpha )(x - \beta )(x - \nu ) = 0$.

Complete step by step answer:
(a) Given, the zeroes of the polynomial,$ - 5 + 2i$ ,$3$ multiplicity $2$. Therefore, we have three zeros of a fourth degree polynomial. We know, if a complex number is a zero of a polynomial with real coefficients, then it’s conjugate is also a zero of the polynomial.So, as $ - 5 + 2i$ is a zero of the polynomial, then, $ - 5 - 2i$ is a zero of the polynomial.So, we have four zeros for a polynomial of fourth degree, that are,
$ - 5 + 2i, - 5 - 2i,3,3$
[Since, $3$ has multiplicity 2]
Therefore, the polynomial is,
$(x - 3)(x - 3)\{ x - ( - 5 + 2i)\} \{ x - ( - 5 - 2i)\} = 0$

Simplifying the expression, we get,
$ \Rightarrow {(x - 3)^2}\{ x + 5 - 2i\} \{ x + 5 + 2i\} = 0$
$ \Rightarrow {(x - 3)^2}\{ (x + 5) - 2i\} \{ (x + 5) + 2i\} = 0$
Using, $(a + b)(a - b) = {a^2} - {b^2}$, we get,
$ \Rightarrow {(x - 3)^2}\{ {(x + 5)^2} - {\left( {2i} \right)^2}\} = 0$
Now, opening the squares, we get,
$ \Rightarrow ({x^2} - 6x + 9)\{ ({x^2} + 10x + 25) - (4{i^2})\} = 0$
We know,${i^2} = - 1$, therefore,
$ \Rightarrow ({x^2} - 6x + 9)\{ ({x^2} + 10x + 25) - ( - 4)\} = 0$

Opening the brackets, we get,
$ \Rightarrow ({x^2} - 6x + 9)\{ {x^2} + 10x + 25 + 4\} = 0$
$ \Rightarrow ({x^2} - 6x + 9)({x^2} + 10x + 29) = 0$
Multiplying the brackets, we get,
$ \Rightarrow {x^2}({x^2} + 10x + 29) - 6x({x^2} + 10x + 29) + 9({x^2} + 10x + 29) = 0$
$ \Rightarrow ({x^4} + 10{x^3} + 29{x^2}) - (6{x^3} + 60{x^2} + 174x) + (9{x^2} + 90x + 261) = 0$
$ \Rightarrow {x^4} + 10{x^3} + 29{x^2} - 6{x^3} - 60{x^2} - 174x + 9{x^2} + 90x + 261 = 0$
Now, simplifying the equation, we get,
$ \therefore {x^4} + 4{x^3} - 22{x^2} - 84x + 261 = 0$

Therefore, the required polynomial is ${x^4} + 4{x^3} - 22{x^2} - 84x + 261 = 0$.

(b) Given the zeros of the polynomial, $ - 4, - i, - 3 + i$. Therefore, we have three zeros of a fifth degree polynomial. We know, if a complex no. is a zero of a polynomial with real coefficients, then it’s conjugate is also a zero of the polynomial. So, $ - i$ is a zero of the polynomial, then, $i$ is also a zero of the polynomial. Similarly, as $ - 3 + i$ is a zero of the polynomial, then, $ - 3 - i$ is also a zero of the polynomial. So, we have five zeros for a polynomial of fifth degree, that are,
$ - 4, - i,i, - 3 + i, - 3 - i$
Therefore, the polynomial is,
$\{ x - ( - 4)\} \{ x - ( - i)\} \{ x - (i)\} \{ x - ( - 3 + i)\} \{ x - ( - 3 - i)\} = 0$

Opening the brackets and simplifying the expression, we get,
$ \Rightarrow \{ x + 4\} \{ x + i\} \{ x - i\} \{ x + 3 - i\} \{ x + 3 + i\} = 0$
$ \Rightarrow \{ x + 4\} \{ x + i\} \{ x - i\} \{ (x + 3) - i\} \{ (x + 3) + i\} = 0$
Using, $(a + b)(a - b) = {a^2} - {b^2}$, we get,
$ \Rightarrow (x + 4)({x^2} - {i^2})\{ {(x + 3)^2} - {i^2}\} = 0$
$ \Rightarrow (x + 4)({x^2} - {i^2})\{ ({x^2} + 6x + 9) - {i^2}\} = 0$
We know that ${i^2} = - 1$, therefore,
$ \Rightarrow (x + 4)\{ {x^2} - ( - 1)\} \{ ({x^2} + 6x + 9) - ( - 1)\} = 0$
$ \Rightarrow (x + 4)\{ {x^2} + 1\} \{ {x^2} + 6x + 9 + 1\} = 0$
Adding up the like terms, we get,
$ \Rightarrow (x + 4)({x^2} + 1)({x^2} + 6x + 10) = 0$

Now, opening the brackets, we get,
$ \Rightarrow (x + 4)\{ {x^2}({x^2} + 6x + 10) + 1({x^2} + 6x + 10)\} = 0$
$ \Rightarrow (x + 4)\{ {x^4} + 6{x^3} + 10{x^2} + {x^2} + 6x + 10\} = 0$
$ \Rightarrow (x + 4)\{ {x^4} + 6{x^3} + 11{x^2} + 6x + 10\} = 0$
Multiplying the brackets with each other, we get,
$ \Rightarrow x({x^4} + 6{x^3} + 11{x^2} + 6x + 10) + 4({x^4} + 6{x^3} + 11{x^2} + 6x + 10) = 0$
$ \Rightarrow ({x^5} + 6{x^4} + 11{x^3} + 6{x^2} + 10x) + (4{x^4} + 24{x^3} + 44{x^2} + 24x + 40) = 0$
Now, simplifying the equation, we get,
$ \therefore {x^5} + 10{x^4} + 35{x^3} + 50{x^2} + 34x + 40 = 0$

Therefore, the required polynomial is ${x^5} + 10{x^4} + 35{x^3} + 50{x^2} + 34x + 40 = 0$.

Note: A polynomial with real coefficients, if it has a complex number as its zero, then the conjugate of the complex number must also be the zero of the polynomial. Complex roots always occur in pairs. We should take care while doing the calculations so as to be sure of the right answer. One should have a good grip over the concepts of polynomials and its roots.