Find the value fo a, if the divisiion of $a x^{3} + 9 x^{2} + 4 x - 10 b y x + 3$ leaves a remainder 5.

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Solution

Let p(x) = $ax^3 + 9x^2 + 4x – 10$

According to the given condition, p(x) leaves the remainder 5 when divided by (x + 3).

⇒ p(–3) = 5

⇒ $a(–3)^3 + 9 (–3)^2 + 4 (–3) – 10$ = 5

⇒ -27a + 81 – 12 – 10 = 5

⇒ -27a = – 54

⇒ a = $\frac{54}{27}$ = 2

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