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Question
If `x^3+ x^2-ax + b` is divisible by `(x^2-x)`,write the value of a and b.
Solution
Equating `x^2 – x `to 0 to find the zeroes, we will get
`x(x – 1) = 0`
`⇒ x = 0 or x – 1 = 0`
`⇒ x = 0 or x = 1`
Since, `x^3 + x^2 – ax + b` is divisible by x`2 – x`.
Hence, the zeroes of `x^2 – x` will satisfy `x^3 + x2 – ax + b`
`∴ (0)^3 + 0^2 – a(0) + b = 0`
`⇒ b = 0`
And
`(1)^3 + 1^2 – a(1) + 0 = 0 [∵b = 0]`
`⇒ a = 2`
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