Question

How many zeros does a quadratic polynomial have?

Answer 1

Hint: First, we need to know about the concept of the quadratic polynomial.
Quadratic equations being the equations that are often called the second-degree equation, like the linear equation is called the first-degree equations Quad is referred to as the square of the terms. In several ways we can calculate the quadratic equation; however, the two methods are most common, which are the standard equation method and factoring method.

Complete step-by-step answer:
The formula for the quadratic equation is generalized as $ ax{}^2 + bx + c = 0 $ where $ a,b,c $ are the numerical coefficients or constant terms from the given and the value x is the unknown and that the value we need to find while solving the quadratic equation.
Let us take the example to show how many zeros in the quadratic polynomial.
Assume the generalized equation be $ ax{}^2 + bx + c = 0 $ is $ x{}^2 + 2x - 15 = 0 $ , that is the value of $ a $ is $ 1 $ , the value of $ b $ is $ 2 $ , and the value of $ c $ is $ - 15 $ ).
Let us solve further, $ x{}^2 + 2x - 15 = 0 \Rightarrow x{}^2 + 5x - 3x - 15 = 0 $ (the two x can be separated to factorizes)
 $ x{}^2 + 5x - 3x - 15 = 0 \Rightarrow x(x + 5) - 3(x + 5) = 0 $ (Taking out the common terms from both sides)
Hence, we get $ x(x + 5) - 3(x + 5) = 0 \Rightarrow (x - 3)(x + 5) = 0 $
 $ (x - 3)(x + 5) = 0 \Rightarrow x - 3 $ or $ x + 5 = 0 $
Turning the values to the right side, we get $ x = 3, - 5 $ (zeroes means the value of x)
Therefore, the zeros of the polynomial have a maximum of $ 2 $ zeroes (if we apply the $ 2 $ zeroes, we get the values of the x)

Additional information:
The above-used form is the factorization method.
The standard method for the quadratic equation is in the form of $ \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} $
In this method, we get the same values as above like, $ x{}^2 + 2x - 15 = 0 $ where (that is the value of $ a $ is $ 1 $ , the value of $ b $ is $ 2 $ , and the value of $ c $ is $ - 15 $ )
Applying the values we get, $ \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} \Rightarrow \dfrac{{ - 2 \pm \sqrt {{2^2} - 4( - 15)} }}{{2(1)}} $
Solving this we get, $ \dfrac{{ - 2 \pm \sqrt {{2^2} - 4( - 15)} }}{2} = \dfrac{{ - 2 \pm \sqrt {4 + 60} }}{2} \Rightarrow \dfrac{{ - 2 \pm 8}}{2} \Rightarrow 3, - 5 $

Note: There is no possibility that $ a $ can never be $ 0 $ .
Suppose $ a = 0 $ in the given generalized polynomial, then we get $ ax{}^2 + bx + c = 0 \Rightarrow bx + c = 0 $ then this equation is the linear equation, as it has a degree $ 1 $ .
And if we solve the linear equation like, $ bx + c = 0 \Rightarrow x - 1 = 0 \Rightarrow x = 1 $ (where $ b $ is $ 1 $ and $ c $ is $ - 1 $ ) we get the zeroes as $ 1 $ . Hence for the linear equation, we have $ 1 $ zeroes.
Thus $ a = 0 $ is not possible in the quadratic equation.