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Question
If both x – 2 and `x - 1/2` are factors of px2 + 5x + r, show that p = r.
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Solution
Let p(x) = px2 + 5x + r
Given (x – 2) is a factor of p(x)
So, p(2) = 0
p(2)2 + 5 × 2 + r = 0
4p + 10 + r = 0 ...(1)
Again, `(x - 1/2)` is a factor of p(x).
So, `p(1/2)` = 0
Now, `p(1/2) = p(1/2)^2 + 5 xx (1/2) + r`
= `1/4p + 5/2 + r`
`p(1/2) = 0`
`1/4 p + 5/2 + r = 0`
From (1), we have 4p + r = –10
From (2), we have p + 10 + 4r = 0
p + 4r = –10
4p + r = p + 4r
3p = 3r
p = r
Hence, proved.
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