If (x+3) divides \( \Large x^{3}+px-81 \) and the remainder is zero, then the value of p is
Correct Answer: Description for Correct answer:
\( \Large f \left(x\right) = x^{3}+px-81 \)
\( \Large \left(x+3\right) is\ a\ factor\ \left(given\right) \)
\( \Large f \left(-3\right) = 0 \)
\( \Large \left(-3\right)^{3}+p \left(-3\right)-81 = 0 \)
\( \Large -27-3p-81 = 0 \)
\( \Large -3p-108 = 0 \)
3p = -108
\( \Large p = \frac{-108}{3} =\ -36 \)
\( \Large f \left(x\right) = x^{3}+px-81 \)
\( \Large \left(x+3\right) is\ a\ factor\ \left(given\right) \)
\( \Large f \left(-3\right) = 0 \)
\( \Large \left(-3\right)^{3}+p \left(-3\right)-81 = 0 \)
\( \Large -27-3p-81 = 0 \)
\( \Large -3p-108 = 0 \)
3p = -108
\( \Large p = \frac{-108}{3} =\ -36 \)
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