Question

if x -3 and x - 1/3 are are factors of the polynomial px2 +3x +r ,show that p=r​

Answer 1

Step-by-step explanation:

Given :

(x - 3) and (x - 1/3) are are factors of the polynomial px² + 3x + r

To prove :

p = r

Solution :

Let q(x) = px² + 3x + r

If (x - a) is a factor of the polynomial q(x), then q(a) = 0

(x - 3) is a factor :

x - 3 = 0

x = 3

Put x = 3 then q(3) = 0,

⇒ p(3)² + 3(3) + r = 0

⇒ 9p + 9 + r = 0

⇒ 9 = -9p - r  --[eqn 1]

(x - 1/3) is a factor :

x - 1/3 = 0

x = 1/3

Put x = 1/3 then q(1/3) = 0,

⇒ p(1/3)² + 3(1/3) + r = 0

⇒ p(1/9) + 1 + r = 0

⇒ p/9 + 1 + r = 0

     

Multiply the above equation by 9,

9(p/9 + 1 + r) = 9(0)

⇒  p + 9 + 9r = 0

⇒  p + 9r + (-9p - r) = 0 [ ∵ eqn [1] ]

⇒  p + 9r - 9p - r = 0

⇒  8r - 8p = 0

⇒  8r = 8p

 ⇒ r = p

Hence proved