The problem 15 is to divide the polynomial $a^2b - ab^2 + \frac{3}{8}a^2b^2$ by $-\frac{5}{3}ab$. The problem 16 is to divide the polynomial $\frac{1}{3}x^4y^3 - \frac{1}{5}x^3y^2 + \frac{1}{4}x^2y$ by $-\frac{1}{5}xy^2$.

1. Problem Description

The problem 15 is to divide the polynomial $a^2b - ab^2 + \frac{3}{8}a^2b^2$ by $-\frac{5}{3}ab$.

The problem 16 is to divide the polynomial $\frac{1}{3}x^4y^3 - \frac{1}{5}x^3y^2 + \frac{1}{4}x^2y$ by $-\frac{1}{5}xy^2$.

2. Solution Steps

Problem 15:

We divide each term of the polynomial by $-\frac{5}{3}ab$.

$\frac{a^2b}{-\frac{5}{3}ab} = a^2b \cdot \frac{-3}{5ab} = \frac{-3a}{5}$

$\frac{-ab^2}{-\frac{5}{3}ab} = -ab^2 \cdot \frac{-3}{5ab} = \frac{3b}{5}$

$\frac{\frac{3}{8}a^2b^2}{-\frac{5}{3}ab} = \frac{3}{8}a^2b^2 \cdot \frac{-3}{5ab} = \frac{-9ab}{40}$

Adding the terms together we get:

$\frac{-3a}{5} + \frac{3b}{5} - \frac{9ab}{40}$

Problem 16:

We divide each term of the polynomial by $-\frac{1}{5}xy^2$.

$\frac{\frac{1}{3}x^4y^3}{-\frac{1}{5}xy^2} = \frac{1}{3}x^4y^3 \cdot \frac{-5}{xy^2} = \frac{-5}{3}x^3y$

$\frac{-\frac{1}{5}x^3y^2}{-\frac{1}{5}xy^2} = -\frac{1}{5}x^3y^2 \cdot \frac{-5}{xy^2} = x^2$

$\frac{\frac{1}{4}x^2y}{-\frac{1}{5}xy^2} = \frac{1}{4}x^2y \cdot \frac{-5}{xy^2} = \frac{-5x}{4y}$

Adding the terms together we get:

$\frac{-5}{3}x^3y + x^2 - \frac{5x}{4y}$

3. Final Answer

For problem 15: $\frac{-3a}{5} + \frac{3b}{5} - \frac{9ab}{40}$

For problem 16: $\frac{-5}{3}x^3y + x^2 - \frac{5x}{4y}$