A rectangle with positive area has length represented by the expression 3x² + 5x - 8 and width by 2x² + 6x. Write expressions in terms of x for the perimeter and area of the rectangle. Give your answers in standard polynomial form and show your work. d. Is it possible for the perimeter of the rectangle to be 16 units? If so, what value(s) of xx will work? Use mathematical reasoning to explain how you know you are correct.

Step-by-step explanation:

P = 2(L + W)

P = 2(3x^2 + 5x - 8 + 2x^2 + 6x)

P = 2(5x^2 +11x - 8)

P = 10x^2 + 22x - 16

Area = L * W

Area = (3x^2 + 5x - 8)(2x^2 + 6x)

Area = 6x^4 + 18x^3 + 10x^3 + 30x^2 - 16x^2 - 48x

Area = 6x^4 + 18x^3 + 40x^3 - 16x^2 - 48x

Part D

The perimeter would have to equal 16. Does it? Could it?

10x^2 + 22x - 16 = 16         Subtract 16 from both sides.

10x^2 + 22x - 32  = 0

x = 1 makes this equation true

x = - 3.2 also makes this equation true.

Ignore the -3.2 for the moment. Lets look at the 1

10(1)^2 + 22(1) - 16

10 + 22 - 16

32 - 16

16

So if x = 1, this perimeter can be 16.

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step-by-step explanation: