Chapter A.3, Problem 62E

Solution

Simplify.

  $\frac{-(2x+5)}{(x+2)(x+3)}$ .

Given:

The given fraction is $\frac{5}{x^2+x-6}-\frac{2}{x-2}+\frac{4}{x^2-4}$ .

Concept Used:

• Remove any grouping symbol such as brackets and parentheses by multiplying factors.
• Use the exponent rule to remove grouping if the terms are containing exponents.
• Combine the like terms by addition or subtraction.
• Combine the constants.

Calculation:

The given fraction is $\frac{5}{x^2+x-6}-\frac{2}{x-2}+\frac{4}{x^2-4}$ .

  $\begin{array}{c}\frac{5}{x^2+x-6}-\frac{2}{x-2}+\frac{4}{x^2-4}=\frac{5}{x^2+x-6}-\frac{2}{x-2}+\frac{4}{x^2-2^2}\\ =\frac{5}{x^2+x-6}-\frac{2}{x-2}+\frac{4}{(x+2)(x-2)}\end{array}$

  $\begin{array}{c}solve{x}^2+x-6=x^2+(3-2)x-6\\ =x^2+3x-2x-6\\ =x(x+3)-2(x+3)\\ =(x-2)(x+3)\\ putthevalueabove\end{array}$

  $\frac{5}{x^2+x-6}-\frac{2}{x-2}+\frac{4}{(x+2)(x-2)}=\frac{5}{(x-2)(x+3)}-\frac{2}{x-2}+\frac{4}{(x+2)(x-2)}$

  $\begin{array}{c}=\frac{5(x+2)-2(x+2)(x+3)+4(x+3)}{(x-2)(x+2)(x+3)}\\ =\frac{5x+10-2(x^2+3x+2x+6)+4x+12}{(x-2)(x+2)(x+3)}\end{array}$

  $\begin{array}{c}=\frac{5x+10-2x^2-6x-4x-12+4x+12}{(x-2)(x+2)(x+3)}\\ =\frac{5x+10-2x^2-6x}{(x-2)(x+2)(x+3)}\end{array}$

  $\begin{array}{c}=\frac{-2x^2-x+10}{(x-2)(x+2)(x+3)}\\ =\frac{-(2x^2+x-10)}{(x-2)(x+2)(x+3)}\end{array}$

  $\begin{array}{c}solve2x^2+x-10=2x^2+(-4+5)x-10\\ =2x^2-4x+5x-10\\ =2x(x-2)+5(x-2)\\ =(2x+5)(x-2)\\ putthevalueabove\end{array}$

  $\begin{array}{c}\frac{5}{x^2+x-6}-\frac{2}{x-2}+\frac{4}{(x+2)(x-2)}=\frac{-(2x^2+x-10)}{(x-2)(x+2)(x+3)}\\ =\frac{-(2x+5)(x-2)}{(x-2)(x+2)(x+3)}\\ =\frac{-(2x+5)}{(x+2)(x+3)}\end{array}$

Thus by simplifying $\frac{5}{x^2+x-6}-\frac{2}{x-2}+\frac{4}{x^2-4}$ to a single fraction it comes $\frac{-(2x+5)}{(x+2)(x+3)}$ .