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\( 3x^{4}\)-\( 2x^{3}\)+\( 3x^{2}\)-\( 2x\)+\( 3\) divided by \( \left( 3x+2\right)\), then the remainder is

A) \( 0\)
B) \( \Large \frac{185}{27} \)
C) \( \Large \frac{181}{25} \)
D) \( \Large \frac{3}{4} \)

Correct Answer:


B) \( \Large \frac{185}{27} \)

Description for Correct answer:
Here

\( \Large f(x) = 3x^{4} - 2x^{3} + 3x^{2} -2x + 3 \)

For 3x + 2 = 0

= > x = -2/3

Now remainder

\( \Large =f \left(- \frac{2}{3} \right) = 3\left(- \frac{2}{3} \right)^{4} -2\left(- \frac{2}{3} \right)^{3} +3\left(- \frac{2}{3} \right)^{2} - 2\left(- \frac{2}{3} \right) + 3 \)

\( \Large = 3 \times \frac{16}{81} + \frac{16}{27} +\frac{12}{9} + \frac{4}{3} + 3 \)

\( \Large = \frac{16}{27} + \frac{16}{27} + \frac{12}{9} + \frac{4}{3} + \frac{3}{1}\)

\( \Large \frac{16 + 16 + 36 + 36 + 81}{27} = \frac{185}{27} \)

Part of solved CDS Maths(2) questions and answers : Exams >> CDSE >> CDS Maths(2)

Comments

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