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Question
The expression 2x3 + ax2 + bx – 2 leaves remainder 7 and 0 when divided by 2x – 3 and x + 2 respectively. Calculate the values of a and b.
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Solution
Let f(x) = 2x3 + ax2 + bx – 2
2x – 3 = 0 `\implies x = 3/2`
On dividing f(x) by 2x – 3, it leaves a remainder 7.
∴ `2(3/2)^3 + a(3/2)^2 + b(3/2) - 2 = 7`
`27/4 + (9a)/4 + (3b)/2 = 9`
`(27 + 9a+ 6b)/4 = 9`
27 + 9a + 6b = 36
9a + 6b – 9 = 0
3a + 2b – 3 = 0 ...(1)
x + 2 = 0 `\implies` x = –2
On dividing f(x) by x + 2, it leaves a remainder 0.
∴ 2(–2)3 + a(–2)2 + b(–2) – 2 = 0
–16 + 4a – 2b – 2 = 0
4a – 2b – 18 = 0 ...(2)
Adding (1) and (2), we get,
7a – 21 = 0
a = 3
Subsituting the value of a in (1), we get,
3(3) + 2b – 3 = 0
9 + 2b – 3 = 0
2b = –6
b = –3
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