Answer: Solved.
Step-by-step explanation: The calculations are as follows -
(12) Given zeroes are 5, -3 and -1 + 3i. So, the other conjugate of -1 + 3i, i.e., -1 - 3i will also be a root of the polynomial. So the polynomial f(x) will be of degree 4 and is given by
[tex]f(x)=(x-5)(x+3)(x+1-3i)(x+1+3i)\\\\\Rightarrow f(x)=(x^2-2x-15)(x^2+2x+1-9i^2)\\\\\Rightarrow f(x)=(x^2-2x-15)(x^2+2x+10)\\\\\Rightarrow f(x)=x^4-9x^2-50x-150.[/tex]
(13) Given polynomial is
[tex]f(x)=x^3-7x^2+9x-24.[/tex]
Now, we will substitute the rational numbers in place of 'x' and check whether the value of f(x) becomes zero or not.
We will see that
[tex]f(-1)\neq 0,~~f(1)\neq 0,~~f(2)\neq 0,~~f(-2)\neq 0, ~~f(3)\neq 0, ~~f(-3)\neq 0,~~etc[/tex]
Also, the polynomial is not zero for any rational number.
(14) Given, [tex]f(x)=7x+6~~\textup{and}~~g(x) = 4x^2.[/tex]
So,
[tex](f+g)(x)=f(x)+g(x)=7x+6+4x^2.[/tex]
Thus, the problems are solved.