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प्रश्न
Find the values of constants a and b when x – 2 and x + 3 both are the factors of expression x3 + ax2 + bx – 12.
उत्तर
Let f(x) = x3 + ax2 + bx – 12
x – 2 = 0
`\implies` x = 2
x – 2 is a factor of f(x).
So, remainder = 0
∴ (2)3 + a(2)2 + b(2) – 12 = 0
`\implies` 8 + 4a + 2b – 12 = 0
`\implies` 4a + 2b – 4 = 0
`\implies` 2a + b – 2 = 0 ...(1)
x + 3 = 0
`\implies` x = –3
x + 3 is a factor of f(x).
So, remainder = 0
∴ (–3)3 + a(–3)2 + b(–3) – 12 = 0
`\implies` –27 + 9a – 3b – 12 = 0
`\implies` 9a – 3b – 39 = 0
`\implies` 3a – b – 13 = 0 ...(2)
Adding (1) and (2), we get,
5a – 15 = 0
`\implies` a = 3
Putting the value of a in (1), we get,
6 + b – 2 = 0
`\implies` b = – 4
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