{(y²-121)(64-x²)}{8y+xy+88+11x} is equal to

Correct Answer is : (y−11)(8−x)(y - 11)(8 - x)(y−11)(8−x)

Correct Answer is : (y−11)(8−x)(y - 11)(8 - x)(y−11)(8−x)

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CTET Paper II (Mathematics and Science) 27 Dec 2021 Solved Paper

Question : 38 of 150

Marks: +1, -0

\frac{(y^{2}-121)(64-x^{2})}{8y+xy+88+11x}

is equal to

Solution:

Concept:

Use the difference of squares formula

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

and factorisation by grouping.

Explanation:

First, factor the numerator.

y^{2} - 121 = (y-11)(y+11)

64 - x^{2} = (8-x)(8+x)

So numerator becomes

(y-11)(y+11)(8-x)(8+x)

Now factor the denominator:

8y+xy+88+11x

Group terms:

(8y+88) + (xy+11x) = 8(y+11) + x(y+11) = (y+11)(8+x)

Cancel the common factors

(y+11)

and

(8+x)

from numerator and denominator.

The expression simplifies to

(y-11)(8-x)

Answer:

Option C:

(y - 11)(8 - x)

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