$15 a b^{2}$ + $2 a b^{2} x$ + $3 a b^{2}$ + $2 a b^{2} x$ + $2 a b^{2}$ = _____

$4 a b^{2}$ + $20 a b^{2} x$

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$24 a^{2} b^{4} x$

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$20 a b^{2}$ + $4 a b^{2} x$

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$24 a b^{2} x$

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Solution

The correct option is C
$20 a b^{2}$ + $4 a b^{2} x$

$15 a b^{2}$ + $2 a b^{2} x$ + $3 a b^{2}$ + $2 a b^{2} x$ + $2 a b^{2}$

= ( $15 a b^{2}$ + $3 a b^{2}$ + $2 a b^{2}$ ) + ( $2 a b^{2} x$ + $2 a b^{2} x$ )

= $(15 + 3 + 2) a b^{2}$ + $(2 + 2) a b^{2} x$

= $20 a b^{2}$ + $4 a b^{2} x$

This cannot be simplified further as $20 a b^{2}$ and $4 a b^{2} x$ are unlike terms.

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