If when $f (x)$ is divided by $3 x + 1$ , the quotient is $x^{2} - x + 3$ and the remainder is $2$ , then find $f (x)$ .

3 x^{3} - 2 x^{2} - 8 x + 3

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3 x^{3} - 2 x^{2} + 8 x + 5

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3 x^{3} + 4 x^{2} + 8 x + 1

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3 x^{3} - 4 x^{2} + 8 x - 1

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3 x^{3} - 4 x^{2} + 8 x - 3

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Solution

The correct option is B $3 x^{3} - 2 x^{2} + 8 x + 5$
Given, Divisor $= 3 x + 1$ , quotient $= x^{2} - x + 3$ and remainder $= 2$
We know, dividend $=$ divisor $\times$ quotient $+$ remainder
$∴ f (x) = (3 x + 1) (x^{2} - x + 3) + 2$
$∴ f (x) = 3 x^{3} + x^{2} - 3 x^{2} - x + 9 x + 3 + 2 = 3 x^{3} - 2 x^{2} + 8 x + 5$

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Similar questions

What must be subtracted from $(4 x^{3} - 3 x + 5)$ to get $(2 x^{2} - 3 x^{3} + 5 x - 2)$ ?

4 x^{3} - 3 x + 5

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

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

(x - 3)

(x + 3)

p (x) = 2 x^{4} - 3 x^{3} - 5 x^{2} + 8 x - 3

p (x)

(x - \frac{3}{2})

\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}

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