If both `x-2` and `x-1/2` are factors of `p x^2+5x+r ,` show that `p=rdot`

Explanation

Let `f(x)=px^2+5x+r`
Now, `g(x)=x-2`
`g(x)=0`
`x-2=0`
`x=2`
Put `x=2` in `p(x)`,
`p(2)=p(2)^2+5(2)+r`
`=4p+10+r`
At `x=2`,
`p(x)=0`
`4p+10+r=0` —-------------- (1)
Now, `h(x)=x-1/2`
Put `x=1/2` in `p(x)`,
`p(1/2) =p(1/2)^2+5(1/2)+r`
`=p/4+5/2+r`
`=(p+2(5)+4r)/4`
`=(p+10+4r)/4`
At `x=1/2`
`p(x)=0`
(p+10+4r)/4=0`
p+10+4r=0` —------------ (2)
On comparing equation (1) and equation (2),
`=>4p+10+r=p+10+4r`
`=>4p-p+10-10=4r-r`
`=>3p=3r`
`=>p=r`
Hence proved.