Question

# Find the value fo a, if the divisiion of ax3+9x2+4x−10byx+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 = 5427 = 2

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