When solving for Financial Polynomials I need to use the formula P (1 + r/2)2. I will be able to calculate how much interest my money will collect over a 1 year period. Then I can further figure out if I will have enough money over a longer period of time, to purchase my new item. I will use $200 at 10% interest for the first equation. The second equation I will use $5,670 at 3.5% interest rate. The final equation I will be dividing -3x by -9×3 + 3×2 – 15x. We need to use the polynomial expression P (1 + r) 2 We will have to eliminate parentheses by multiplying by itself P (1 + r) (1 + r) Using foil to carry out the expressionP (1 + r + r + r2) Combine like termsP (1 + 2r +r2)
Distribute the P in the trinomial P + 2Pr + Pr2 Normal polynomials run in descending order however; this one runs in ascending order with my highest exponent as the last term instead of first term. Page 304 problem 90 of Elementary and Intermediate Algebra states “P dollars is invested at annual rate r for 1 year. If the interest is compounded seminally than the polynomial is P (1 + r) 2 represents the value of the investment after 1 year. Rewrite the expression without parentheses. Evaluate the expression using P = 200 and r = 10%.” First the r is turned into a decimalP = 200 r = .10
My formula after removing parenthesesP + 2Pr + Pr2 Substitute my values200 + 2(200) (.10) + 200(.10)2 After doing my exponent and math on both sides 200 + 40 + 2 The answer of my formula is 242 Over the course of 1 year, if I invested $200 at 10% interest rate. I would receive $42 in interest for that year. My total amount save thus far is $242. The next set I will be using is P = $5670 r = 3.5%
Changing my 3.5% into a decimalr = .035 Expanded formula to useP + 2Pr + Pr2 Substituting my values given5670 + 2(5670) (.035) + 5670(.035)2 After doing my exponent and math on both sides5670 + 396.9 + 6.95 Answer to formula is 6073.85 Over the course of 1 year, if I invested $5,670 at 3.5% interest rate. I would receive $403.85 in interest. The total amount saved after 1year would be $6,073.85.
These calculations are important in the real world because when you want to save money for something big over time. You need to know your options of interest rates and how much you need to invest to have the money you will need over time. Page 311 problem 70 of Elementary and Intermediate Algebra states “-9×3 + 3×2 – 15x / -3x.” My expression to start (-9×3 + 3×2 – 15x) / (-3x)
After rewriting to know what is divided-9×3 + 3×2 – 15x -3x Place my divisor under all dividends-9×3 + 3×2 – 15x -3x -3x -3x Dived all variables subtract all exponents final answer is 3×2 – x +5 The interest rate of 10% on my invested $200, gave me $42 extra at the end of the year. The interest rate of 3.5% on my invested $5,670, gave me $403.85 extra at the end of the year. If you plug your new amounts into the equation again and have the same interest rate. Your new interest you would receive on the $242 will be $50.82 for a total of $292.82 in the second year. Your interest on my new amount of $6073.85 will be $432.61 for a total of $6506.46 in the second year.
When you divide expressions you can skip the second part. When you know the expression you are doing simply divide your divisor by all your dividends there is no need to write it down; unless you are showing your work in full detail. After I have the numbers I will go back and subtract all my exponents from each other. For a final answer with less work involved. The problem would have gone like this: -9×3 + 3×2 -15x / (-3x) Answer would have been quick and simple 3×2 –x – 5