Find quadratic polynomials whose zeros are -3,3 ?

Hint: For the above problem first we will write the generalized quadratic equation in order to find the equation with the given zeros(factors).
The general quadratic equation is given as;
$a{x^2} + bx + c$ (x is the variable, a, b and c are the constants)
Using the above generalized quadratic equation we will find the given with the two zeros.

Complete step-by-step solution:
In order to find the equation with given zeros, before proceeding further first we have to learn a few more concepts;
Suppose, the two zeros of the generalized equation are $\alpha ,\beta $.
From the generalized equation$a{x^2} + bx + c$ we must infer that the product of two zeros is equal to;
$\alpha \beta = \dfrac{c}{a}$ and the sum of two zeros is equal to $\alpha + \beta = \dfrac{{ - b}}{a}$ .
We will apply the rules for calculating the quadratic equation;
Zeros given to us -3 and 3, therefore product of two zeros is;
$ \Rightarrow - 3 \times 3 = \dfrac{c}{a}$ (multiplied the two given zeros)
$ \Rightarrow - 9 = \dfrac{c}{a}$ (term is negative)
$ \Rightarrow - 9a = c$
Sum of the two zeros is;
$ \Rightarrow - 3 + 3 = \dfrac{{ - b}}{a}$
$ \Rightarrow 0 = \dfrac{{ - b}}{a}$ (We got the zero sum of the two factors given to us)
We got the product and sum now we can form the equation by substituting the sum and product in general equation;
$ \Rightarrow a{x^2} - 9a = 0$ (We will take a common out)
$ \Rightarrow {x^2} - 9 = 0$
$ \Rightarrow {x^2} - {(3)^2} = 0$ is the required equation.

Therefore ${x^2} - {(3)^2} = 0$ is the required answer.

Note: In order to solve the quadratic equation we have two methods one is the splitting the middle term and the other is the standard method having the formula for finding its zeros, the formula is stated as follows; $\dfrac{{b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$ , plus and minus sign indicates it will give values of two factors of any given quadratic equation.