You have already seen that the zero of a linear polynomial ax + b is – b/a. We will now try to explore the relationship between zeroes and coefficients of a quadratic polynomial. For this, let us take the quadratic polynomial p(x) = 2x²– 8x + 6.

    In Class-IX, we have learnt how to factorise quadratic polynomials by splitting the middle term. So, we split the middle term '-8x' as a sum of two terms, whose product is 6 x 2x²- = 12x². So, we write

2x² 8x + 6 - 2x² - 6x - 2x + 6

= 2x(x-3) - 2(x-3)

= (2x-2)(x-3) - 2(x-1) (x-3)

  p(x) = 2x² - 8x + 6 is zero when x-1 = 0 or x-3 = 0, ie., when x-1 or x=3. So, the zeroes of 2x²-8x+6 are 1 and 3. We now try and see if these zeroes have some relationship to the coefficients of terms in the polynomial. The coefficient of x² is 2; of x is-8 and the constant is 6, which is the coefficient of x⁰. (ie. 6x⁰ = 6)

We see that the sum of the zeroes-1+3-4= -(-8)/2=(coefficient of x)/coefficient of x²

Product of the zeroes 1×3=3=6/2=constant term/2 coefficient of x²

Let us take one more quadratic polynomial

p(x)=3x² + 5x - 2.

By splitting the middle term we see,

3x²+5x-2 = 3xM² + 6x - x - 2- 3x (x+2)-1(x+2)

    =(3x-1)(x+2)

3x² + 5x - 2 is zero when either 3x-1-0 or x+2=0

ie, when x=1/3 or x=-2.

The zeroes of 3x² + 5x-2 are and-2. We can see that the:

Sum of its zeroes = 1/3+(-2)=-5/3=-(coefficient of x)/ coefficient of x²

Product of its zeroes = 1/3×(-2)=-2/3=constant term/coefficient of x²