Factorize: $a^{2} b^{2} + 8 a b - 9$

(a b - 1) (a b + 9)

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(a b - 9) (a b + 1)

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(a b + 1) (a b + 9)

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(a b - 9) (a b + 9)

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Solution

The correct option is A $(a b - 1) (a b + 9)$
$\Rightarrow a^{2} b^{2} + 8 a b - 9$

$Split (8) into two parts such that$
$their product is -9 and their sum is 8$

$\Rightarrow a^{2} b^{2} - 1 a b + 9 a b - 9$

$\Rightarrow a b (a b - 1) + 9 (a b - 1)$

$\Rightarrow (a b - 1) (a b + 9)$

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