## Financial Polynomials Essay

Financial PolynomialsAshford UniversityAbstractIn this paper I will be demonstrating how to use financial polynomials with a few expressions from the textbook “Elementary and Intermediate Algebra”. I will not only show the problem, but also will also break the expression down showing all mathematical work, and provide reasoning of how anybody can apply this theory to everyday life. In the paper there will be the following words: FOIL, like terms, Descending order, Dividend, and Divisor highlighted and explained. In the text we are given the following expression .

With this expression we are to evaluate the polynomial using : P=\$200 and r= 10%, andP=\$5670 and r=3.5%First we have to rewrite the expression without the parenthesis. One way to do this is to use a process called the FOIL method were we will multiple across the binomial using the steps of the FOIL method. P(1+r/2)2 ?The original expressionP(1+r/2)*(1+r/2) ? Square the quantity (1+r/2)2 this will cancel out the exponent P(1+r/2+r/2+r2/2) ? Here is were the FOIL method comes into play when there are Like terms they need to be combined.

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P(1+r+r2/4)? After FOIL method move (P) across the expression P+Pr+Pr2/4? After parenthesis are moved here is what our expression looks like. Now we are ready to move to our second step inputting the above numbers into our expression. P+Pr+Pr2/4? The original expression(P)200+(P)200*((r )0.1)+(p)200*((r) 0.1)24I’ve arranged all the integers into their rightful space in the expression. This is the order we will solve the expression. All values at this point should be substituted into the formula (P)200+(P)200*((r )0.1)+(p)200*((r) 0.

1)2 = 200+20+200*0.01=2 4 4 200+20=220+2/4 We can reduce 2/4 to lowest terms which is ½ so the expression now reads 220+1/2 = 220.50. With a interest of 10% and \$200 invested in one years time this account would have made \$20.50 in interest.Our second item we are to solve is the same expression just with different values. Since we already have our expression without parenthesis, then there is no reason to rework the expression again.

Here are our values: P=\$5670r=3.5%P+Pr+Pr2/4 ? Our expression5670+5670*(0.035) ? We cannot put 3.5% into our expression therefore we must it into its decimal form which is 0.035. 5670+198.

45? Multiply 0.035 by 5670 to receive 198.45 5670+198.45=\$5868.

45? we will put this number to the side for right now and work the second half of the expression. Earlier in our expression we had 0.035, which was equals r2. 0.0352 = 0.001225.

5670*0.001225=6.94575 With a dividend of 6.94575 into the divisor of 4= 1.7364375 4 4\$5868.

45+1.7364375=\$5,870 ? Bring down the \$5868.45 that we solved for earlier in the expression and add it to the 1.7364375, which bring us to our answer.

\$5870 invest with 3.5% interest will earn \$200 within a year.Our last and final exercise is to divide polynomial by a monomial. Here is our expression: (-9 x3+3 x2 -15 x)/(-3 x)The way I found it easiest to solve this expression was to break it down into sections starting from left to right descending order down the expression.

-9×3+ 3×2 – 15x-3x -3x -3x This process makes it a little easier to understand. From here take the -3x anddivide it across the expression leaving you with an answer of 3×2-x+5x+1. Youcan check this answer by just plugin in -3x across the new express to arrive at the first expression in the exercise.ConclusionIn this brief paper I have demonstrated the use of polynomials to solve simple compound interest expressions. I briefly introduced terms such as: FOIL, like terms, descending order, dividend, and divisor. These steps demonstrated here today can be used in everyday accounting.

Weather you are planning a vacation get away in a couple years or saving for retirement this formula will help you figure out how much you need to invest with a given interest to reach your goals. I hope this short essay was informative and gave you a better understanding of financial polynomials.BIBLIOGRAPHYdugopolski, m.

(2009). Exponents and Polynomials. Elementary and Intermediate Algebra 4th edition (p.

304). new york: McGraw Hill.

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