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Question
Find the value of a, if the division of ax3 + 9x2 + 4x – 10 by x + 3 leaves a remainder 5.
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Solution
Let f(x) = ax3 + 9x2 + 4x – 10
x + 3 = 0
`\implies` x = –3
On dividing f(x) by x + 3, it leaves a remainder 5.
∴ f(–3) = 5
a(–3)3 + 9(–3)2 + 4(–3) – 10 = 5
–27a + 81 – 12 – 10 = 5
-27a + 81 - 22 = 5
-27a + 59 = 5
-27a = 5 - 59
-27a = -54
a = `(-54)/-27`
a = 2
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