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Question
If x3 + ax2 + bx + 6 has x – 2 as a factor and leaves a remainder 3 when divided by x – 3, find the values of a and b.
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Solution
Let f(x) = x3 + ax2 + bx + 6
∴ x – 2 = 0 `\implies` x = 2
(2)3 + a(2)2 + b(2) + 6 = 0
8 + 4a + 2b + 6 = 0
4a + 2b + 14 = 0
2(2a + b + 7) = 0
2a + b + 7 = `0/2`
2a + b + 7 = 0
2a + b = –7 ...(i)
∴ x – 3 = 0 `\implies` x = 3
(3)3 + a(3)2 + b(3) + 6 = 3
27 + 9a + 3b + 6 = 3
9a + 3b + 33 = 3
9a + 3b = 3 – 33
9a + 3b = –30
3(3a + b) = –30
3a + b = `(-30)/3`
3a + b = –10 ...(ii)
Subtracting (i) from (ii), we get,
2a + b = – 7
3a + b = – 10
– – +
– a = 3
∴ a = –3
Substituting the value of a = –3 in (i), we get,
2a + b = –7
2(–3) + b = –7
– 6 + b + 7 = 0
b = –7 + 6
∴ b = –1
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