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

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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 - 4 This 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