: A quadratic polynomial, the sum of whose zeros is 0 and one zero is 3, is
A.${{x}^{2}}-9$
B.${{x}^{2}}+9$
C.${{x}^{2}}+3$
D.${{x}^{2}}-3$

Hint: For solving this problem, first express the general quadratic equation in terms of sum of zeroes and product of zeros. Let the other zero be k. Now the product of zeros can be obtained by using the sum of zeros condition. Putting both the values in the general form will give us the desired result.

Complete step-by-step answer:
In algebra, a quadratic function is a polynomial function with one or more variables in which the highest-degree term is of the second degree. A single-variable quadratic function can be stated as:
$f(x)=a{{x}^{2}}+bx+c,\quad a\ne 0$
If we have two zeros of a quadratic equation then the polynomial could be formed by using the simplified result which could be stated as:
${{x}^{2}}-(a+b)x+ab$, where a and b are two zeroes of the equation.
According to the problem statement, we are given that the sum of zeros of a quadratic polynomial is 0. Putting it in the general form, we get
$\begin{align}
  & a+b=0 \\
 & \therefore {{x}^{2}}-0x+ab=0 \\
 & {{x}^{2}}+ab=0 \\
\end{align}$
Let the other zero be k. Since the sum of zeros is 0, therefore k + 3 = 0. So, the other zero is -3
The product of zeros can be expressed as: $3\times -3=-9$
Putting the obtained value in the general form, we get
$\begin{align}
  & ab=-9 \\
 & {{x}^{2}}+ab={{x}^{2}}-9 \\
\end{align}$
Hence, the required quadratic polynomial is ${{x}^{2}}-9$.
Therefore option (a) is correct.

Note: After obtaining the second root of the equation, we can directly evaluate the quadratic equation as $(x-3)(x+3)={{x}^{2}}-9$, without using the general form. Since, it is mentioned that the polynomial is a quadratic polynomial, so we can proceed in this manner also to save time.