If the area of the rectangle is $9 x^{2} - 12 x - 32$ , then the perimeter of the rectangle is .

2 (3 x - 10) (3 x + 2)

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2 (3 x + 10) (3 x - 2)

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2 (6 x - 4)

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Solution

The correct option is C $2 (6 x - 4)$
Given,
The area of the rectangle $= 9 x^{2} - 12 x - 32$

Now,
$9 x^{2} - 12 x - 32$ can be written as
$= (3 x)^{2} - 2 \times 3 x \times 2 + 4 - 36 = (3 x - 2)^{2} - 6^{2} = (3 x - 2 + 6) \times (3 x - 2 - 6) = (3 x + 4) \times (3 x - 8)$
[Using the identity, $(a + b) \times (a - b) = a^{2} - b^{2}$ ]

Now,
Length $= 3 x + 4$ ; Breadth $= 3 x - 8$
Perimeter $= 2 \times (l e n g t h + b r e a d t h) = 2 \times (3 x + 4 + 3 x - 8) = 2 \times (6 x - 4)$ )

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