Question

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

Answer 1

px*2 + 5x + r is a quadratic function. If (x-2) and (x-1/2) are its factors, then they are all the factors there are up to a multiplicative constant, i.e., you can express px*2 + 5x + r as c(x-2)(x-1/2), where c is some constant. Now you can compare coefficients of x*2 and the constant terms of px*2 + 5x + r and c(x-2)(x-1/2) = cx*2 - 2.5cx + c to deduce that p = c and r = c, which of course means that p = r as required.

Answer 2

Given,
(x-2) & (x-1/2) are factors
So,
P(x)=2
p(2^2)+5×2+r=0
4p+r+10=0-----------eq.1

P(x)=1/2
p(1/2)^2+r+5/2=0
p/4+r+5/2=0----------eq.2
Multiply eq.2 with 4
p+4r+10=0-------------eq.3
Now, eq.1-eq.3
You will get
3p-3r=0
3p=3r
Finally on, p=r