Q. 2 I3.7( 4 Votes )

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Answer :

In given expression


Take out the common factor,


[3 × 3 × 3 × 3 × x × x × x × x - 11 × 11 × x × x]


x × x[3 × 3 × 3 × 3 × x × x - 11 × 11]


x2[81x2 – 121]


In expression 81x2 - 121


Both terms are perfect square


81x2 = 9x × 9x


121 = 11 × 11


81x2 – 121 Seems to be in identity a2-b2 = (a + b)(a-b)


Where a = 9x and b = 11;


81x2 – 121 = (9x-11)(9x + 11)


Hence the factors of 81x4 – 121x2 are x2,(9x-11) and (9x + 11)


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