Example : Find the zeroes of the quadratic polynomial ${x^2 + 2x - 35}$, and verify the relationship between the zeroes and the coefficients.

Solution 
			$x^2  + 2x - 35 = 0$
			$ x^2  + 7x - 5x - 35=0$
			$ x(x+7) -5(x+7)=0$
			$ (x+7)(x-5)=0$
			$ x= -7 $ and $5$ 
			
	Therefore the zeroes of the polynomial $p(x)$ are -7 and 5.
	
	coefficient of ${x^2}$ =  1
   coefficient of ${x}$   =   2
   Constant term     =  -35

Sum of its zeroes  = α +  β = ${-b/a}$  = ${-Coefficient of x}/{coefficient of x^2}$
                   = (-7)+ (5)= -2 = ${-2}/{1}$

Product of its zeroes = αβ  = ${c/a}$ = ${Constant term}/{coefficient of x^2}$
                      = (-7) x (5) = 35 = ${-35}/{1}$
	

Example : Find the zeroes of the quadratic polynomial ${x^2 + 7x + 10}$, and verify the relationship between the zeroes and the coefficients.

Solution : 
             ${x^2  + 7x + 10 = (x + 2)(x + 5)}$
  So, the value of ${x^2  + 7x + 10}$ is zero when ${x + 2 = 0}$ or ${x + 5}$ = 0, i.e., when ${x = – 2}$ or ${x = –5}$. 
  Therefore, the zeroes of ${x^2  + 7x + 10}$ are – 2 and – 5. 
  coefficient of ${x^2}$ =  10
  coefficient of ${x}$   =   7
  Constant term     =  10

Sum of its zeroes  = α +  β = ${-b/a}$  = ${-Coefficient of x}/{coefficient of x^2}$
                   = (-2)+ (-5)= -7 = ${-7}/{1}$

Product of its zeroes = αβ  = ${c/a}$ = ${Constant term}/{coefficient of x^2}$
                      = (-2) x (-5) = 10 = ${10}/{1}$

					  

Example: Find the zeroes of the polynomial ${x^2 – 3}$ and verify the relationship between the zeroes and the coefficients.

Solution : 
          Recall the identity ${a^2  – b^2 }$ = ${(a – b)(a + b)}$. Using it, we can write: 
          ${x^2  – 3}$ =  ${(x-√3)(x+√3)}$
	So, the value of ${x^2  – 3}$ is zero when ${x}$ = √3 or -√3 

  coefficient of ${x^2}$ = 1 
  coefficient of ${x}$   = 0  
  Constant term     =  -3

Sum of its zeroes  = α +  β = ${-b/a}$  = ${-Coefficient of x}/{coefficient of x^2}$
                   = ${(√3)+ (-√3)}$= 0 = ${0}/{1}$ = 0

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

Example: Find a quadratic polynomial, whoes zeroes are $1/2$ and $-5$, respectively.

Solution: Let the quadratic polynomial be ${ax^2  + bx + c}$, and its zeroes be α and β. 
		Given, α = $1/2$ and β = $-5$
		
		Sum of zeroes = α + β =  $1/2 -5$ = $-9/2$ = $-b/a$
		Product of zeroes = αβ = $1/2  × (-5)$ = $-5/2$ = $c/a$
		
		So ${ax^2  + bx + c}$ = ${x^2  + b/ax + c/a}$
					= $x^2  + 9/2x + (-5/2)$
					= $2x^2  + 9x -5$
Ans:So, the required polynomial is $2x^2  + 9x -5$ 

Example: Find a quadratic polynomial, the sum and product of whose zeroes are – 3 and 2, respectively.

Solution : 
          Let the quadratic polynomial be ${ax^2  + bx + c}$, and its zeroes be α and β. We have
			α + β =  – 3 =   ${-b/a}$ 
			and αβ =  2 = ${c/a}$

If a = 1, then b = 3 and c = 2.
So, one quadratic polynomial which fits the given conditions is ${x^2  + 3x + 2}$.

Example: Find a quadratic polynomial, the sum and product of whose zeroes are 8 and -65, respectively.

Solution : 
          Let the quadratic polynomial be ${ax^2  + bx + c}$, and its zeroes be α and β. We have
			α + β =  8 =   ${-b/a}$ 
			and αβ =  -65 = ${c/a}$

If a = 1, then b = -8 and c = -65.
So, one quadratic polynomial which fits the given conditions is ${x^2  - 8x -65}$.

Example: Find a quadratic polynomial, the sum and product of whose zeroes are $5/6$ and $1/6$, respectively.

Solution : 
          Let the quadratic polynomial be ${ax^2  + bx + c}$, and its zeroes be α and β. We have
			α + β =  $5/6$ =   ${-b/a}$ 
			and αβ =  $1/6$ = ${c/a}$

If a = 1, then b = -$5/6$ and c = $1/6$.
${ax^2  + bx + c}$ => ${x^2  + (-5/6)x + 1/6}$
So, one quadratic polynomial which fits the given conditions is ${6x^2  - 5x +1}$.