Chapter A.3, Problem 23E

Solution

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

  $(x+3)(x+4)$ .

Given:

The given rational expression is $\frac{x+3}{x-2}=\frac{?}{x^2+2x-8}$ .

Calculation:

The given rational expression is $\frac{x+3}{x-2}=\frac{?}{x^2+2x-8}$ .

As given the two rational expressions are equal.

  $\begin{array}{c}\frac{x+3}{x-2}=\frac{?}{x^2+2x-8}\\ \frac{(x+3)(x^2+2x-8)}{x-2}=\frac{?}{1}\\ solve{x}^2+2x-8=x^2+(4-2)x-8\\ =x^2+4x-2x-8\\ =x(x+4)-2(x+4)\\ =(x-2)(x+4)\\ putthevalueabove\\ \frac{(x+3)(x^2+2x-8)}{x-2}=\frac{?}{1}\\ \frac{(x+3)(x-2)(x+4)}{(x-2)}=\frac{?}{1}\\ \frac{(x+3)(x+4)}{1}=\frac{?}{1}\\ (x+3)(x+4)=?\end{array}$

So the rational expression will be $\frac{x+3}{x-2}=\frac{(x+3)(x+4)}{x^2+2x-8}$ .

Thus the missing numerator (?) is = $(x+3)(x+4)$ .