Chapter 4.2, Problem 53E

(a)

To determine

To find-A polynomial for the value of person B’s account on January 1, 2015.

Explanation of Solution

Given information:

Person A’s account worth on January 1, 2015= $2000{(1+r)}^3+3000{(1+r)}^2+1000(1+r)$

Consider the table for Person A and B:

  

So, the polynomial expression for B’s account on January 1, 2015 will be = $5000{(1+r)}^3+1000{(1+r)}^2+4000(1+r)$

(b)

To determine

To calculate: The total value of the two accounts as a polynomial

Explanation of Solution

Given information:

Person A’s account worth on January 1, 2015= $2000{(1+r)}^3+3000{(1+r)}^2+1000(1+r)$

Calculation:

  

So, the polynomial expression for B’s account on January 1, 2015 will be = $5000{(1+r)}^3+1000{(1+r)}^2+4000(1+r)$

So, the total value of the polynomial expression for A and B’s account is = $2000{(1+r)}^3+3000{(1+r)}^2+1000(1+r)$ + $5000{(1+r)}^3+1000{(1+r)}^2+4000(1+r)$ = $7000{(1+r)}^3+4000{(1+r)}^2+5000(1+r)$

(c)

To determine

To calculate: The total value of the two accounts on January 1, 2015

Explanation of Solution

Given information:

Person A’s account worth on January 1, 2015= $2000{(1+r)}^3+3000{(1+r)}^2+1000(1+r)$

Interest rate = $0.05$

Calculation:

  

So, the polynomial expression for B’s account on January 1, 2015 will be = $5000{(1+r)}^3+1000{(1+r)}^2+4000(1+r)$

So, the total value of the polynomial expression for A and B’s account is = $2000{(1+r)}^3+3000{(1+r)}^2+1000(1+r)$ + $5000{(1+r)}^3+1000{(1+r)}^2+4000(1+r)$ = $7000{(1+r)}^3+4000{(1+r)}^2+5000(1+r)$

So, total values of the two accounts on January 1, 2015 = $7000{(1+0.05)}^3+4000{(1+0.05)}^2+5000(1+0.05)=17763.375$