Assignment 38
"Life is only good for two things; discovering mathematics and teaching mathematics."  (Simeon Poisson)
Caroline Herschel (17501848) and Mary Somerville were named as the first honorary female members of the Royal Astronomical Society. Herschel and her brother, William, are considered by some to be the founders of modern astronomy. Although Caroline was a equal partner in their many accomplishments  such as discovering many comets and nebulae  William received the lion's share of the credit, including being listed as the discoverer of the planet Uranus. Women were simply not considered equals in fields such as math and science. Caroline's honorary membership in the Royal Astronomical Society was bestowed upon her when she was 85 years old.

Why did Herkimer think that tired people from the East would never make it to California?
Answer: He heard that there was no west for the weary.
Herky's friends:
EDDIE KETT ... this guy was always very polite.
TIM BURR ... a lumberjack. 
ASSIGNMENT #38
Reading: Section 6.4, pages 345347.
Written: Page 349/6884 (evens), 87,88,89. In the group 6884, solve as many as you can by factoring. 
Mathematical word analysis: GOOGOL: The number 1 followed by 100 zeros; that is, 10^{100}. The name was created by the nine year old nephew of Dr. Edward Kasner. Interestingly, the largest reasonable estimate for the number of particles in the universe is 10^{85}. A googol is 10^{15} = 1,000,000,000,000,000 times this number. 
First off, let me say that many financial problems involve the solving of polynomials. Here is a very simple example: If $100 grows to $120 in two years, what is the annual interest rate. OK, let X be the unknown rate. Then we must solve
$100(1+X)^{2} = $120
==> (1+X)^{2} = 1.2 ==> 1+X = ÷(1.2) ==> X = ÷(1.2)  1 = 0.095445.
The annual rate is about 9.54%.
Note that (1+X)^{2} is a polynomial. In this problem there is no need to write it as X^{2} + 2X + 1.
Example: Solve 2x^{3}  6x^{2} = 0.
Solution: 2x^{3 } 6x^{2} = 0
==> 2x^{2}(x  3) = 0
==> x = 0 or x = 3.
Example: Solve x^{4} 13x^{2} +36 = 0
Solution: x^{4} 13x^{2} +36 = 0
==> (x^{2}  4)(x^{2}  9) = 0
==> x^{2} = 4 or x^{2} = 9
==> x = 2 or x = 2 or x = 3 or x = 3.
The solution set is {3, 2, 2, 3}.
Problem: Solve x^{4 }+ 7x^{3}  8x  56 = 0.
Solution (with communication):
One could produce a calculator solution, but this complicatedlooking equation can be solved by factoring.
x^{4 }+ 7x^{3}  8x  56 = 0
==> x^{3}(x+7)  8(x+7) = 0
==> (x+7)(x^{3}8) = 0
==> x = 7 or x^{3} = 8
==> x = 7 or x = 2.
Check: Both numbers make the original conditional statement true.

Problem: The width of the base of a rectangular box must be 2 feet longer than its length, and the box must be 4 feet high. Find the dimensions if the box must be designed to hold 21 cubic feet.
Solution (with communication):
Let x = length of base (in feet).
Then x+2 is the width measure (in feet).
We must have 4x(x+2) = 21
==> 4x^{2} + 8x  21 = 0
==> (2x3)(2x+7) = 0
==> 2x = 3 or 2x = 7
==> x = 1.5 or x = 3.5.
Since the domain of x is x > 0, we conclude that x = 1.5 is the only meaningful solution. The dimensions of the box are 1.5 ft. by 3.5 ft. by 4 ft.
Check: (1.5)(3.5)(4) = 21. 
