Q. 1 F5.0( 3 Votes )

# Factories the following expression-

81x^{2}– 198 xy + 121y^{2}

Answer :

In the given expression

1^{st} and last terms are perfect square

⇒ 81x^{2} = 9x × 9x

⇒ 121y^{2} = 11y × 11y

And the middle expression is in form of 2ab

198xy = 2 × 9x × 11y

∴ 9x × 9x - 2 × 11y × 9x + 11y × 11y

Gives (a-b)^{2} = a^{2}-2ab + b^{2}

⇒ In 81x^{2} – 198xy + 121y^{2}

a = 9x and b = 11y;

∴ 81x^{2} – 198xy + 121y^{2} = (9x-11y)^{2}

Hence the factors of 81x^{2} – 198xy + 121y^{2} are (9x-11y)and (9x-11y)

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