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Question
x3 − 23x2 + 142x − 120
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Solution
Let `f(x) = x^3 - 23x^2 + 142x -120` be the given polynomial.
Now, putting x=1,we get
`f(1) = (1)^3 - 23(1)^2 + 142(1) - 120`
` = 1 -23 + 142 - 120`
` = 143 - 143 = 0`
Therefore, (x-1)is a factor of polynomial f(x).
Now,
`f(x) = x^2(x-1) - 22x(x-1) + 120(x -1)`
`=(x-1){x^2 - 22x + 120}`
` = (x -1) {x^2 + 12x - 10x + 120}`
`=(x - 1)(x - 10)(x-12)`
Hence (x-1),(x-10) and (x-12) are the factors of polynomial f(x).
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