Chapter A.3, Problem 25E

Solution

Find the missing numerator or denominator so that the two rational expressions are equal.

  $x(x^2-2x)$ .

Given:

The given rational expression is $\frac{x^2-3x}{?}=\frac{x-3}{x^2-2x}$ .

Calculation:

The given rational expression is $\frac{x^2-3x}{?}=\frac{x-3}{x^2-2x}$ .

As given the two rational expressions are equal.

  $\begin{array}{c}\frac{x^2-3x}{?}=\frac{x-3}{x^2-2x}\\ \frac{(x^2-3x)(x^2-2x)}{?}=\frac{x-3}{1}\\ \frac{x(x-3)(x^2-2x)}{?}=\frac{x-3}{1}\\ \frac{x(x^2-2x)}{?}=\frac{x-3}{(x-3)}\\ \frac{x(x^2-2x)}{?}=1\\ x(x^2-2x)=?\end{array}$

So the rational expression will be $\frac{x^2-3x}{x(x^2-2x)}=\frac{x-3}{x^2-2x}$ .

Thus the missing denominator (?) is = $x(x^2-2x)$ .