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

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

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

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

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

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Solution

The correct option is D $(a b - 1) (a b + 9)$
$a^{2} b^{2} + 8 a b - 9$
$= a^{2} b^{2} + 9 a b - a b - 9$
$= a b (a b + 9) - (a b + 9)$
$= (a b + 9) (a b - 1)$

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Similar questions

a^{2} b^{2} + 8 a b - 9

a^{2} b^{2} + 8 a b - 9

a^{2} b^{2} + 8 a b - 9

∣ ∣ ∣ ∣ \begin{matrix} 1 & a & b - a & 1 & c - b & - c & 1 \end{matrix} ∣ ∣ ∣ ∣ =

a b > - 1, b c > - 1

c a > - 1

{cot}^{- 1} (\frac{a b + 1}{a - b}) + {cot}^{- 1} (\frac{b c + 1}{b - c}) + {cot}^{- 1} (\frac{c a + 1}{c - a})

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