If the area of a rectangle is 6x^2-5x-4 and if the width is 2x+1 then what would the length be

Given: width = x - 1
P = 4x + 6
What is area, A?
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We need to again determine the length of the other side (in this case, the Length), in terms of x

P = 2 w + 2 L
P = 2 (x - 1) + 2 ( L) = 4x + 6
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subtract 2x - 2 from both sides:
2 (L) = 4x + 6 - (2x -2)
2 (L) = 2x + 8
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divide both sides by 2:
L = x + 4
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Now we know both sides:
w = x - 1 (given)
L = x + 4 (calculated)
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To determine the Area, A, we just multiply the two sides together:
A = (x - 1) (x + 4)
A = x^2 - 1x + 4x - 4
A = X^2 + 3x - 4This is actually two problems. 
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First one:
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Let A = area, which is short side (width) x long side (length)
Given, A = 6x^2 + 5x + 1
Given one side = 3x + 1 (we don't know if this the width or length)
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If A = side x side, and we are given A and one side, we can divide A by the given side to derive the remaining (unknown) side:
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Since A is in the form of a quadratic equation, if we factor it, the two factors are the equations in terms of x for each of the two sides:
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(3x + 1) (?x + ??) = 6x^2 + 5x + 1
(3x + 1) (2x + 1) are the two factors and thus, are the two sides. The missing side we needed to solve for is:
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2x + 1
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However, we are asked for its perimeter in terms of x. The equation for P, the perimeter, is:
P = 2 x width + 2 x length
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We already determined the width = 2x + 1
We were given length = 3x + 1
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P = 2 (2x + 1) + 2 (3x + 1)
P = 10x + 4