Chapter 7.4, Problem 20Q

a.

To determine

The polynomial in the standard form that represents the balance in the account after two years.

Explanation of Solution

Given information:

  $\$1000$ Is deposited in an account that earns interest compounded annually.

The expression that shows the balance after two years is,

  $\begin{array}{l}$ Amount=1000{(1+r)}^2$r=\text{rate of interest in decimal form compounded annually}\text{.}\end{array}$

Calculations:

Here the expression that the amount after two years is,

  $\begin{array}{l}$ Amount=1000{(1+r)}^2$r=\text{rate of interest in decimal form compounded annually}\text{.}\end{array}$

Simplifying the expression we get,

  $\begin{array}{l}$ 1000{(1+r)}^2$1000(1+r^2+2r)\\ \Rightarrow 1000+1000r^2+2000r\end{array}$

Thus the polynomial that shows the amount after two years is,

  $1000+1000r^2+2000r$

Now we write the polynomial in standard form according to the descending order of degree of the terms in the polynomial.

  $1000r^2+2000r+1000$

Thus the standard form of the polynomial is $1000r^2+2000r+1000$

b.

To determine

The balance in the account after two years if the rate of interest is $3\%.$

Explanation of Solution

Given information:

  $\$1000$ Is deposited in an account that earns interest compounded annually.

The expression that shows the balance after two years is,

  $\begin{array}{l}$ Amount=1000{(1+r)}^2$r=\text{rate of interest in decimal form compounded annually}\text{.}\end{array}$

Calculations:

Here the expression that the amount after two years is,

  $\begin{array}{l}$ Amount=1000{(1+r)}^2$r=\text{rate of interest in decimal form compounded annually}\text{.}\end{array}$

Simplifying the expression we get,

  $\begin{array}{l}$ 1000{(1+r)}^2$1000(1+r^2+2r)\\ \Rightarrow 1000+1000r^2+2000r\end{array}$

Thus the polynomial that shows the amount after two years is,

  $1000+1000r^2+2000r$

Now the rate of interest is $r=3\%=0.03\text{ (in decimal form)}$

Now putting the value of rate of interest I the expression, we get

  $\begin{array}{l}Amount=1000+1000r^2+2000r\\ $ Amount=1000+1000{(0.03)}^2+2000(0.03)$Amount=1000+0.09+30=1030.9\end{array}$

Therefore the balance in the account after two years is $\$1030.9$ .

c.

To determine

If there is enough money in the account to buy the guitar after three years.

Explanation of Solution

Given information:

  $\$1000$ Is deposited in an account that earns interest compounded annually.

The expression that shows the balance after two years is,

  $\begin{array}{l}$ Amount=1000{(1+r)}^2$r=\text{rate of interest in decimal form compounded annually}\text{.}\end{array}$

The cost of the guitar is $\$1100$

Calculations:

Here the expression that the amount after two years is,

  $\begin{array}{l}$ Amount=1000{(1+r)}^2$r=\text{rate of interest in decimal form compounded annually}\text{.}\end{array}$

The expression for the amount after three years is,

  $\begin{array}{l}$ Amount=1000{(1+r)}^2$r=\text{rate of interest in decimal form compounded annually}\text{.}\end{array}$

Now the rate of interest is $r=3\%=0.03\text{ (in decimal form)}$

Now putting the value of rate of interest I the expression, we get

  $\begin{array}{}$ Amount=1000{(1+r)}^3$$ Amount=1000{(1+0.03)}^3$$ Amount=1000{(1.03)}^3$$ Amount=1000\times 1.092727$$ Amount=1092.727$\end{array}$

Therefore the balance in the account after three years is $\$1092.727$ .

So there is not enough balance in the account after three years to buy the guitar.