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Question
If both x − 2 and \[x - \frac{1}{2}\] are factors of px2 + 5x + r, then
Options
p = r
p + r = 0
2p + r = 0
p + 2r = 0
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Solution
As (x - 2)and (x - 1/2)are the factors of the polynomial `px^2 + 5x + r`
i.e., f(2) = 0and `f(1/2) = 0`
Now,
`f(2) = p(2)^2 + 5(2) + r = 0`
`4p + r = -10 ..... (1)`
And
`f(1/2) = p(1/2)^2 + 5(1/2) + r = 0`
`p/4 + 5/2 + r = 0`
`p + 10 + 4x = 0`
`p+ 4x = -10 ........(2)`
From equation (1) and (2), we get
`4p + r = p + 4r`
`3p = 3x`
` p = r`
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