Sum and Product of zeros in a Quadratic Polynomial
2026-02-28 23:56 Diff
https://brightchamps.com/en-us/math/algebra/sum-and-product-of-zeroes-in-a-quadratic-polynomial

In this section, we will discuss the relation between the coefficients and the sum and product of the zeros of the polynomial. Let the roots of a quadratic polynomial in the form f(x) = ax2 + bx + c be α and β. If the roots are α and β, then the sum is: α + β = -b/a. Their product is: αβ = c/a.

For example, finding the sum and product of zeros in f(x) = 2x2 + 5x + 3
Here, a = 2, b = 5, and c = 3.

Using the formula to find the sum and product of the roots:
α + β = -b/a and αβ = c/a
α + β = -5/2 
αβ = 3/2

Let’s verify the answer by finding the roots of the quadratic polynomial. To find the roots, we use the quadratic formula:
x = -b ± b2 - 4ac2a
x = -5 ± 52 - 4 × 2 × 32 × 2
x = -5 ± 25 - (4 × 2 × 3)4
x = -5 ± 25 - 244
x = -5 ± 14 
x = -5 +  14  or x = -5 -  14 
x = -4/4 = -1 or
x = -6/4 = -3/2 
So, x = -1 and x = -3/2
The sum of the roots = -1 + (-3/2) = -5/2 
The product of the roots = -1 × (-3/2) = 3/2
Therefore, the sum and product of the roots are directly related to the coefficients. 

Let’s understand how the sum and product of a quadratic polynomial’s roots are related to its coefficients. The factored form of a quadratic polynomial is: f(x) =  (x - a)(x - b), where a and b are the roots. Let the sum of the roots be S = a + b and the product be P = ab. 
Expanding the equation: (x - a)(x - b) = x2 - ax - bx + ab
= x2 - (a + b)x + ab
So, P(x) = x2 - Sx + P. 
Here, the coefficient of x is -S, so it is the negative of the sum of the roots and the term P is the product of the roots. 

Now, let’s generalize the result for any quadratic polynomial of the form: f(x) = ax2 + bx + c. where ‘a’ is the coefficient of x2, b is the coefficient of x, and c is the constant. Consider the root as m and n, then the polynomial can be factorized as: f(x) = a(x - m)(x - n)

Let’s compare two forms of a quadratic polynomial:  
ax2 + bx + c and a(x - m)(x - n)
Now divide both sides by ‘a’ to simplify:
x2 + (b/a)x + c/a = (x - m)(x - n)
Next, expand the right-hand side: (x - m)(x - n) = x2 - (m + n)x + mn
So now we have:
x2 + (b/a)x + c/a = x2 - (m + n)x + mn
Since both expressions represent the same polynomial, we can match the coefficients:
The coefficient of x: 
b/a = -(m + n)  m + n = -b/a
The constant term:
c/a = mn
So we’ve shown:
Sum of the roots: m + n = -b/a