$must be added to (x^{3} + 3 x - 8) to get$
$(3 x^{3} + x^{2} + 6)$ .

2 x^{3} + 4 x^{2} - 3 x + 14

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2 x^{3} + x^{2} - 3 x + 17

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2 x^{3} + x^{2} - 3 x + 14

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2 x^{3} + x^{2} - x + 14

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Solution

The correct option is C $2 x^{3} + x^{2} - 3 x + 14$
Let the expression to be added be $Y$ .
$(x^{3} + 3 x - 8) + Y = (3 x^{3} + x^{2} + 6)$

$To get 3 x^{3}, we need to add 2 x^{3} as first term to x^{3} in (x^{3} + 3 x - 8)$

$Similarly, as we don't have x^{2} i n (x^{3} + 3 x - 8) . We need to add x^{2} as second term$

$Similarly, we have an extra 3x in (x^{3} + 3 x - 8) . So, we need to add -3x. as third term$

$Similarly, to make -8 to +6 we need add 6 + 8 = 14 to (x^{3} + 3 x - 8) . So, we need to add 14 as fourth term$

$So we have to add (2 x^{3} + x^{2} - 3 x + 14) t o (x^{3} + 3 x - 8), to get the expression (3 x^{3} + x^{2} + 6) .$

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Similar questions

What must be added to $(x^{3} + 3 x - 8)$ to get $(3 x^{3} + x^{2} + 6)$ ?

x^{3} + 3 x - 8

3 x^{3} + x^{2} + 6

What must be added to (x^{3} + 3 x - 8) to obtain (3 x^{3} + x^{2} + 6) ?

x^{2} + 3 x - 8

x^{2} + x + 6

\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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