If the quotient $= 3 x^{2} - 2 x + 1,$ remainder $= 2 x - 5$ and the divisor $= x + 2$ , then the dividend is

3 x^{3} - 4 x^{2} + x - 3

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3 x^{3} - 4 x^{2} - x + 3

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3 x^{3} + 4 x^{2} - x + 3

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3 x^{3} + 4 x^{2} - x - 3

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Solution

The correct option is D $3 x^{3} + 4 x^{2} - x - 3$
Dividend $=$ Divisor $\times$ Quotient $+$ Remainder.
Here, Divisor $\times$ Quotient
$= (x + 2) (3 x^{2} - 2 x + 1) = 3 x^{3} + 6 x^{2} - 2 x^{2} - 4 x + x + 2 = 3 x^{3} + 4 x^{2} - 3 x + 2$
Adding the remainder,
$= 3 x^{3} + 4 x^{2} - 3 x + 2 + 2 x - 5 = 3 x^{3} + 4 x^{2} - x - 3$
$=$ Dividend

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Similar questions

p (x) = 3 x^{3} - 4 x^{2} + 7 x + 5

(x - 3)

(x + 3)

3 x^{3} - 4 x^{2} + 7 x - 5

x = 3

x = - 3

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x^{2} + x \sqrt{x} - 2

x^{4} + 2 x

4 x^{2} + 3 x - 5

\sqrt{2 x} + \sqrt{3}

\begin{matrix} {lim}_{x \to 1} \frac{x^{2} - 3 x + 2}{x^{2} - 4 x + 3} {lim}_{x \to 2} \frac{x^{3} + x^{2} - 12}{x^{3} - x^{2} - x - 2} {lim}_{x \to \sqrt{2}} \frac{x^{2} + x \sqrt{2} - 4}{x^{2} - 3 x \sqrt{2} + 4} {lim}_{x \to 3} \frac{x^{3} - 3 x^{2} - x + 3}{x^{3} - 6 x^{2} + 9 x} \end{matrix}

(3 x^{3}) \times (4 x^{2}) + 7 x^{5}

x = 3

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