Question

21. Paul has $900 to invest in a savings account that has an annual interest rate of 1.8%, and a money market account that pays 4.2% per year. Write a polynomial for the interest he will earn in one year if he invests x dollars in the savings account.

Answer 1

The polynomial that gives the interest earned after a year will have variables, exponents and constants that are joined by operators.

• The interest earned after one year is 0.018·x

Reasons:

The amount Paul has to invest = $900

The annual interest rate from the savings account = 1.8%

The amount the money market account pays per year = 4.2 %

Required: The polynomial for the interest Paul earned by investing x dollars in the savings account.

Solution:

The interest earned is found using the compound interest formula as follows;

$\displaystyle A = \mathbf{P \cdot \left(1 + \frac{r}{n} \right)^{n \cdot t}}$

Where;

A = The amount in the account after one year

P = The original amount invested = x

r = The interest rate offered on the investment = 1.8% = 0.018

t = The time of the investment = 1 year

n = The number of times of application of the interest per period = Once per year

Which gives;

Interest = Amount earned = A - P

Therefore;

$\displaystyle Interest, \ I = \mathbf{P \cdot \left(1 + \frac{r}{n} \right)^{n \cdot t} - P}$

Plugging in the values gives;

$\displaystyle I = x \cdot \left(1 + \frac{0.018}{1} \right)^{1 \times 1} - x = x \cdot 1.018^1 - x = 1.018 \cdot x - x = 0.018 \cdot x$

The polynomial equation is therefore;

Interest, I = 0.018·x

Using the simple interest formula, we have;

$\displaystyle Interest = \mathbf{\frac{P \times r \times t}{100}}$

Which gives;

$\displaystyle Interest = \frac{x \times 1.8 \times 1}{100} = 0.018 \cdot x$

Interest earned by investing in the savings account for one year, I = 0.018·x

• The polynomial representing the interest earned is I = 0.018·x

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