The area of a rectangle is $12 y^{4} + 28 y^{3} - 5 y^{2}$ . If its length is $6 y^{3} - y^{2}$ , then its width is

y + 5

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- 2 y + 5

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- 2 y^{2} + 5

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2 y + 5

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Solution

The correct option is D $2 y + 5$
Area of the rectangle $= 12 y^{4} + 28 y^{3} - 5 y^{2}$
Length of rectangle $= 6 y^{3} - y^{2}$
Let the breadth of the rectangle be $a$
Thus,
$∴ a \times (6 y^{3} - y^{2}) = 12 y^{4} + 28 y^{3} - 5 y^{2}$
$=> a \times y^{2} (6 y - 1) = y^{2} (12 y^{2} + 28 y - 5)$
$=> a \times y^{2} (6 y - 1) = y^{2} (12 y^{2} - 2 y + 30 y - 5)$
$=> a \times y^{2} (6 y - 1) = y^{2} [2 y (6 y - 1) + 5 (6 y - 1)]$
$=> a \times y^{2} (6 y - 1) = y^{2} (6 y - 1) (2 y + 5)$
$=> a = 2 y + 5$
Thus, breadth is $2 y + 5$

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The area of a rectangle is 12y4+28y3−5y2. If its length is 6y3−y2, then its width is

The area of a rectangle is $12 y^{4} + 28 y^{3} - 5 y^{2}$ . If its length is $6 y^{3} - y^{2}$ , then its width is