Consider a quadratic polynomial, $p(x) = 2x^2 – 8x + 6$ . Factorise quadratic polynomials by splitting the middle term. So, split the middle term ‘– 8x’ as a sum of two terms, whose product is $6 × 2x^2$ = $12x^2$.

 Therefore,
		$2x^2 –  8x  +  6$ = $2x^2 – 2x -6x  +  6$
					=$2x(x-1) - 6(x-1)$
					=$(x-1)(2x-6)$
					
So, the value of $p(x)  =  2x^2 –  8x  +  6$ is zero when $x – 1 = 0$ or $2x – 6 = 0$, 
i.e., when $x = 1$ or $x = 3$. 

So, the zeroes of $2x^2 –  8x  +  6$ are 1 and 3. 

Observe that :

Sum of its zeroes  = 1+3=4 = -${8/2}$ = ${-Coefficient of x}/{coefficient of x^2}$

Product of its zeroes = $3×1 =3 $= ${6/2}$ = ${Constant term}/{coefficient of x^2}$

Let us consider a quadratic polynomial, say, $p(x) = 3x^2  + 5x – 2$. 
By the method of splitting the middle term,
$3x^2  + 5x – 2 = 3x^2 +6x - x - 2 $
				=$3x(x+2) -1(x+2)$
				=$(x+2)(3x-1)$

Hence, the value of $3x^2  + 5x – 2$ is zero when either $3x – 1 = 0$ or $x + 2 = 0$, i.e., 

when x = ${1/3}$ or x = –2. So, the zeroes of $3x^2  + 5x – 2$ are ${1/3}$ and – 2. Observe that :

Sum of its zeroes  = ${1/3}-$ 2 = ${-5/3}$  = ${-Coefficient of x}/{coefficient of x^2}$

Product of its zeroes = ${1/3}$x(-2) =${-2/3}$  = ${Constant term}/{coefficient of x^2}$

In general, if α* and β* are the zeroes of the quadratic polynomial p${(x)}$ = a${x^2}$ + b${x}$ + c,
a ≠ 0, then you know that $x – α$ and $x – β$ are the factors of p${(x)}$. Therefore,
a${x^2}$ + b${x}$ + c =  k(${x}$ – α) (${x}$ – β), where k is a constant
=  k[${x^2}$  – (α + β)${x}$ + α β]
=  k${x^2}$  – k(α  + β)${x}$ + k α  β
Comparing the coefficients of ${x^2}$, ${x}$ and constant terms on both the sides, we get
a = k, b =  – k(α  + β) and c = kαβ.

This gives                           α +  β = ${-b/a}$ ,

                                         αβ  = ${c/a}$
										 
Sum of its zeroes  = α +  β = ${-b/a}$  = ${-Coefficient of x}/{coefficient of x^2}$

Product of its zeroes = αβ  = ${c/a}$ = ${Constant term}/{coefficient of x^2}$