Which ratio represents the area of the smaller rectangle compared to the area of the larger rectangle? (Figure not drawn to scale) A rectangle with the dimensions x by x plus two is inside a larger rectangle with the dimenstions two x by x squared plus six x plus eight. x over x plus four x plus two all over x squared plus six x plus eight one over two times the sum of x and four two x times the sum of x and four

The answer is 1/(2(x+4)) I'll type out how I got that

First you multiply the lengths of each rectangle to get the areas x(x+2) 2x(x^2+6x+8) Then to take the ratio, you divide the two: [x(x+2)]/[2x(x^2+6x+8)]

It reduces to (x+2)/[2(x^2+6x+8)] Then you factor the bottom to give: (2+x)/[2(x+4)(x+2)] The top cancels out with one of the bottom terms: 1/[2(x+4)]

Small rectangle x , x +2 dimensions Area Smaller rectangle= x(x+2) Large rectangle 2x , x^2 +6x +8 dimensions Area of larger rectangle = 2x *(x^2+6x+8) Factors of x^2 +6x +8 =(x+4)(x+2) Ratio of smaller to larger area = [x(x+2)]/ [ 2x*(x+2)(x+4)] 1: 2(x+4) One over two times the sum of x and 4

what if it was like x(x+3) 3x(x^2+7x+12) @bme89 @rital1975 @missMob

@Bcoxy8 first expand \[x ^{2} + 7x +12\] into \[(x+3)(x+4)\] That will give you the ratio of \[\frac{ x(x+3) }{3x(x+4)(x+3)} \] then you can eliminate the x and (x+3) to give you...\[\frac{ 1 }{ 3(x+4) }\]

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