Please note that, in terms of eNotes rules, only one question (and one closely related question can be answered at any given time. Please therefore repost your remaining questions as 2 separate posts.

A (sq) = `l^2`

A(rect) = l x b

`therefore A=(x+1)^2` (square) and

`A=(2x+5) times (2x-1)` (retangle)

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Please note that, in terms of eNotes rules, only one question (and one closely related question can be answered at any given time. Please therefore repost your remaining questions as 2 separate posts.

A (sq) = `l^2`

A(rect) = l x b

`therefore A=(x+1)^2` (square) and

`A=(2x+5) times (2x-1)` (retangle)

But the rectangle is also 3 times the area of the square

`therefore A = 3(x+1)^2` (rectangle)

`therefore 3(x+1)^2 = (2x+5)(2x-1)` (these both represent the area of the rectangle). Now simplify:

`therefore 3(x^2 +2x+1 )= 4x^2-2x+10x -5`

`3x^2 +6x +3 = 4x^2 +8x - 5`

Bring everything to the one side:

`therefore 0 = 4x^2-3x^2+8x -6x -5-3`

`0= x^2+2x-8` which can also be written:

`x^2 +2x -8=0`

Now factorize. Use the factors of `1x^2` and the factors of -8 That is `1x times 1x` and for the 8 1x8 would not work but 4x2 will work. Care to arrange the symbols correctly. You can check your work if you are ever unsure :

`(x-2)(x+4)=0`

Each factor is equal to zero

`therefore x-2=0 and x+4=0`

`therefore` **x=2 and x=-4**