SOLUTION: A rectangle with positive area has length represented by the expression 3x^2+5x-8 and width by 2x^2+6x. Give your answers in standard polynomial form. d) Is it possible for the

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: A rectangle with positive area has length represented by the expression 3x^2+5x-8 and width by 2x^2+6x. Give your answers in standard polynomial form. d) Is it possible for the       Log On


   

Question 826696: A rectangle with positive area has length represented by the expression 3x^2+5x-8 and width by 2x^2+6x. Give your answers in standard polynomial form.
d) Is it possible for the perimeter of the rectangle to be 16 units? If so, what value(s) of x will work? Use mathematical reasoning to explain how you know you are correct.
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
d) The perimeter of any polygon is the sum of all the sides. For a rectangle it is 2 times the length plus 2 times the width:
P+=+2%283x%5E2%2B5x-8%29%2B2%282x%5E2%2B6x%29
Simplifying...
P+=+6x%5E2%2B10x-16%2B4x%5E2%2B12x%29
P+=+10x%5E2%2B22x-16
If the perimeter could be 16 then
16+=+10x%5E2%2B22x-16
should have a solution. So we will look for solutions. This is a quadratic so we want one side to be zero. Subtracting 16 from each side:
0+=+10x%5E2%2B22x-32
Now we factor. First the greatest common factor of 2:
0+=+2%285x%5E2%2B11x-16%29
Next we factor the trinomial. (If you have a lot of trouble with this factoring, it is a quadratic so you can use the quadratic formula instead.)
0+=+2%285x%2B16%29%28x-1%29
From the Zero Product Property:
2 = 0 or 5x+16 = 0 or x-1 = 0
Solving these we find that the first equation is impossible so there are no solutions to it. For the other two we should get:
x+=+-16%2F5 or x+=+1

Before we decide that these are the answers, we must first decide if it is OK for x to have these values. We cannot use a value for x that makes the length or width of the rectangle to be negative or zero. So we have to check each solution to see if they make the length and width positive:
Checking x+=+-16%2F5
Length: Check!
Width: Check!

Checking x = 1:
Length: 3x%5E2%2B5x-8+=+3%281%29%5E2%2B5%281%29-8+=+3%281%29+%2B+5%281%29+-8+=+3+%2B+5+-+8+=+0
Failure! The length cannot be zero!

So the answer to part b is: Yes, the rectangle can have a perimeter can be 16. But only if x+=+-16%2F5

P.S. Note that it was the negative value for x that resulted in a valid length and width while the positive value for x did not.