Adog pen in the shape of a rectangle is to be built using one wall of a house as a side. if 132 feet

of fencing is to be used and the pen is to provide the maximum room for the dog, how long

should each side be?

now try to find as many different lengths of fence as you can (other than 132 feet) that will give

whole number pen sides. for each different fence length, give the lengths of each side that give

the maximum room for the dog.

This is a very poorly written question. im assuming the house wall is at least 130 ft since there is no information given about it.

each set of side lengths has three elements, the long side and the two short sides. we're asked to list all whole number possibilities so start anywhere, the two short sides being 1ft
(1 1 130). the long side has to be the difference of 132 and the length of both short sides.

the area is one of the short sides times the ling side, in the first case 1*130=130 sqft

continuing the list

(2 2 128) area=256
(3 3 126) area=378
(4 4 124) area=496
(5 5 122) area=610

the short sides increase by 1, the long side decreases by 2, the area gets bigger, but the difference between the areas are 4 less each time. dividing the first difference, 122, by 4 will give us a number to make a better guess from.

122÷4=30.5

so lets try

(30 30 72) area=2160

lets try 31 and see if that gives a smaller area

(31 31 70) area = 2170

(32 32 68) area =2176

(33 33 66) area = 2178

(34 34 64) area = 2176

so the area is decreasing now which means the maximum area is found when thw long side is 66 feet and the short sides are 33 feet.

as far as listing all possibilities, you see the pattern i started, i leave the rest to you

Answer to your question: a. the fundamental theorem of algebra tells you that the equation will have two complex roots since the degree of the polynomial is 2. the roots are ​ x=1±i7√2 .number of roots=degree of polynomialhere, degree of polynomial is 2 so number of roots will be 2.hope it *_*