Find the quadratic polynomial whose zeroes are $ 3 $ and $ - 5? $

Hint: Here we are given two zeroes of the given polynomial. Zeroes are also known as the roots. First of all we will assume both the given zeros as alpha and beta then will place its value in the standard formula that is $ {x^2} + (\alpha + \beta )x + \alpha \beta = 0 $

Complete step by step solution:
Let us assume that the given zeros are –
 $ \alpha = 3 $ and $ \beta = - 5 $
Now, place the above values in the equation-
 $ {x^2} + (\alpha + \beta )x + \alpha \beta = 0 $
 $ {x^2} + [3 + ( - 5)]x + (3)( - 5) = 0 $
Simplify the above expression –
When you combine plus and minus together then it results in minus, since the product of negative and positive gives negative as the resultant sign.
 $ {x^2} + [3 - 5]x + ( - 15) = 0 $
When you subtract the bigger term from the positive smaller term then the resultant value will be negative after subtraction.
 $ {x^2} + ( - 2)x + ( - 15) = 0 $
Again, applying the same concept product of negative and positive terms gives resultant value as the negative.
 $ {x^2} - 2x - 15 = 0 $
This is the required solution.
So, the correct answer is “ $ {x^2} - 2x - 15 = 0 $ ”.

Note: Be careful about the sign convention when you combine the terms with the different positive and the negative terms. When you combine terms with two different signs you have to do subtraction and give a sign of a bigger term to the resultant value. In multiplication when you multiply terms with different signs then the resultant value would be negative but when you combine two terms with the same sign then the resultant value would be positive.