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

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

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

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Solution

The correct option is B
$2 x^{3} + x^{2} - 3 x + 14$

Let the expression to be added be $y$ .

$(x^{3} + 3 x - 8) + y = 3 x^{3} + x^{2} + 6 \Rightarrow y = (3 x^{3} + x^{2} + 6) - (x^{3} + 3 x - 8)$

$3 x^{3} + x^{2} + 6$
$x^{3} + 3 x - 8$
$- - + - --------------- -$
$2 x^{3} + x^{2} - 3 x + 14$

So we have to add $(2 x^{3} + x^{2} - 3 x + 14)$ to $(x^{3} + 3 x - 8)$ , 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 obtain (3 x^{3} + x^{2} + 6) ?

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

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

x^{2} + 5 x - 6

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

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

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