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Question
Find the values of m and n so that x – 1 and x + 2 both are factors of x3 + (3m + 1)x2 + nx – 18.
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Solution
Let f(x) = x3 + (3m + 1)x2 + nx – 18
x – 1 = 0 `\implies` x = 1
x – 1 is a factor of f(x).
So, remainder = 0
∴ (1)3 + (3m + 1)(1)2 + n(1) – 18 = 0
`\implies` 1 + 3m + 1 + n – 18 = 0
`\implies` 3m + n – 16 = 0 ...(1)
x + 2 = 0 `\implies` x = –2
x + 2 is a factor of f(x).
So, remainder = 0
∴ (–2)3 + (3m + 1)(–2)2 + n(–2) – 18 = 0
`\implies` –8 + 12m + 4 – 2n – 18 = 0
`\implies` 12m – 2n – 22 = 0
`\implies` 6m – n – 11 = 0 ...(2)
Adding (1) and (2), we get,
9m – 27 = 0
m = 3
Putting the value of m in (1), we get,
3(3) + n – 16 = 0
9 + n – 16 = 0
n = 7
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