What should be added to $x^{2} + 3 x - 8$ to get $x^{2} + x + 6$ ? $[1 Mark]$

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Solution

Here we have to subtract
( $x^{2}$ + 3 $x$ - 8) from ( $x^{2}$ + $x$ + 6),

$(x^{2} + x + 6)$
$- (x^{2} + 3 x - 8)$
$(-) (-) (+) - ------------ -$
$0 x^{2} - 2 x + 14$

So, $(- 2 x + 14)$ should be added to $(x^{2} + 3 x - 8)$ to get $(x^{2} + x + 6)$ .

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Similar questions

x^{2} - 3 x + 8

2 x^{2} + x - 6

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

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

x^{3}

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

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