Question

If both x-2 and x-1/2 are factors of px²+5x+r show that p=r

Answer 1

Given, f(x) = px^2+5x+r  and factors are x-2, x-1/2

Substitute x = 2 in place of equation, we get

= p*2^2+5*2+r=0  

=  4p + 10 + r = 0   ------------  (i)

Substitute x = 1/2 in place of equation.  

 p/4 + 5/2 + r = 0

 p + 10 + 4r = 0  -------------------  (ii)

On solving (i),(ii) we get

4p+r=-10 and p+4r+10=0

4p+r=p+4r

3p=3r   

p = r.

Hope this helps!

Answer 2

As both are the factors of the given polynomials, both must be zero for x = 2 and 1/2.   Using factor theorem:

If x - 2 is factor: f(2) = 0

⇒ p(2)² + 5(2) + r = 0

⇒ 4p + 10 + r = 0          ...(1)

If x - 1/2 is factor: f(1/2) = 0

⇒ p(1/2)² + 5(1/2) + r = 0  

⇒ p/4 + 5/2 + r = 0  

⇒ p + 10 + 4r = 0         ...(2)

  Subtract (1) from (2), we get:

⇒ 3p - 3r = 0

⇒ 3p = 3r

⇒ p = r           proved