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Financial Polynomials
Solution for the problem: (-9x3 + 3x2 -- 15x) Bring down the last term, 15x, and divide again by the divisor 3x. The quotient is 5. Again, 5 is placed above the division bracket and added to the final quotient:
The division process for the polynomial above can be approached in the same way as dividing whole numbers. The polynomial (-9x3 + 3x2 -- 15x) is the dividend, while (-3x) is the divisor. To easily facilitate the division process, the whole equation will be multiplied by "-1." The new equation is: (9x3 -- 3x2 + 15x) / (3x). Writing the question in long division form, begin dividing (9x3) first by (3x), which is equal to (3x2). To cancel the first part of the equation, the first part of the quotient must be negative. Thus, 3x2 becomes (-3x2). Place (-3x2) above the division bracket as shown below.
) 9x3-3x2 + 15x
Multiply (3x) by (-3x2), which is equal to 9x3. Placing 9x3 below (-9x3) then subtract them, resulting to zero. The remaining parts of the equation must be divided in a descending order. Bring down the next term, (-3x2), and divide again by the divisor 3x. Dividing (-3x2) by 3x will give the quotient x. This is placed above the division bracket and added to the equation, resulting to the yet unfinished quotient:
____- 3x2 + x
3x
) 9x3-3x2 + 15x
____- 3x2 + x_- 5
3x
) 9x3-3x2 + 15x
-9x3
- 3x2
15x
0
The complete quotient is (-3x2 + x - 5). However, this equation must be multiplied again by "-1" to return the equation terms to their original negative or positive signs since initially the whole equation was also multiplied by "-1" for the problem to be solved easily. Thus, the final solution to the problem is the equation: 3x2 -- x + 5.
Problem 2: A. Solve the polynomial P ( 1 + r) 2.
2
B. Then evaluate the resulting polynomial using the following values:
B1. P = $200 and r = 10%
B2. P = $5,670 and r = 3.5%
Solution to Problem 2:
A. The polynomial P ( 1 + r/2 )2 will be solved by expanding the polynomial first.
P (1 + r/2)2 = P (1 + r/2) (1 + r/2)
Then, using the…
There are many other variables that would affect real-world riding speed, and the effort variable would also be far more complicated than represented here, but this should suffice for now. Several equations can be written using the variables defined here. For instance, to calculate the effort needed to go one kilometer (it's easier to go kilometers than miles, at least mathematically), or a thousand meters, in a given gear,
Susan K. Peterson and associates (1988) conducted a study on the impact of the use of manipulatives on different kinds of students and concluded that the result of using manipulative was positive for both gifted and disabled students (Peterson, Mercer & O'Shea, 1988). Joseph Martinez (1987) also explained that the use of solid manipulatives made studying math more fun as well as less hectic and demanding for most of
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