Question

Compound Interest After 3 years, an investment of \(\$ 1000\) earning an interest rate \(r\) compounded annually will be worth \(1000(1+r)^{3}\) dollars. Write this expression as a polynomial in standard form.

Short answer

The expression \(1000(1+r)^{3}\) as a polynomial in standard form is \(1000r^{3}+3000r^{2}+3000r+1000\).

Step-by-step solution

Expand the expression

The expression can be expanded using the formula for the cube of a binomial, which is \(a^{3}+3a^{2}b+3ab^{2}+b^{3}\). Applying the formula to the given \(1000(1+r)^{3}\), it becomes \(1000(1^3+3\times1^2\times r+3\times1\times r^2+r^3)\) or \(1000(1+3r+3r^2+r^3)\).

Multiply the terms

Multiply every term within parentheses by \(1000\). So the expression transforms into \(1000+3000r+3000r^{2}+1000r^{3}\)

Rearrange in descending powers of \(r\)

Finally, it is needed to arrange the terms in descending power of \(r\). So, the expression in polynomial form becomes \(1000r^{3}+3000r^{2}+3000r+1000\)