Factorise : 121 x^{2}-289 y^{2} | vedclass

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Question

(A) The given expression is $121 x^{2}-289 y^{2}$.We know that $121 = 11^{2}$ and $289 = 17^{2}$.So,the expression can be written as $(11 x)^{2}-(17 y

Vedclass · Accepted Answer

$(11x - 17y)(11x + 17y)$

Vedclass · Answer

(A) The given expression is $121 x^{2}-289 y^{2}$.We know that $121 = 11^{2}$ and $289 = 17^{2}$.So,the expression can be written as $(11 x)^{2}-(17 y)^{2}$.Using the algebraic identity $a^{2}-b^{2}=(a-b)(a+b)$,where $a=11x$ and $b=17y$,we get:$(11 x)^{2}-(17 y)^{2}=(11 x-17 y)(11 x+17 y)$.

Factorise $: 121 x^{2}-289 y^{2}$

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