Chapter 1, Problem 60RE

To determine

To simplify $\frac{1}{x+2}+\frac{1}{x^2-4}+\frac{2}{x^2-x-2}$ by using appropriate operations.

Answer to Problem 60RE

The answer is $\frac{2x^2+2x-2}{x{(x-2)}^2}$ .

Explanation of Solution

Given information:

  $\frac{1}{x+2}+\frac{1}{x^2-4}+\frac{2}{x^2-x-2}$

Formula used:

Properties of fractions:- $\frac{A}{C}+\frac{B}{C}=\frac{A+B}{C}$

Difference of squares : $A^2-B^2=(A-B)(A+B)$

Factoring Trinomials:- To factor the form $x^2+bx+c$

  $(x+r)(x+s)=x^2(r+s)x+rs$ where $r+s=b$ and $rs=c$ .

Calculations:

First Term $\frac{1}{x+2}$

Second Term

  $\begin{array}{l}=\frac{1}{x^2-4}\\ =\frac{1}{(x+2)(x-2)}\end{array}$

Now,

  $\begin{array}{l}=\frac{1}{(x+2)}+\frac{1}{(x+2)(x-2)}\\ =\frac{1}{(x+2)}\left\{1+\frac{1}{(x-2)}\right\}\end{array}$ ( Taking $\frac{1}{(x-2)}$ common)

writing the fraction in LCD (least common denominator) form we have,

  $\begin{array}{l}=\frac{1}{(x-2)}\left\{\frac{x-2+1}{x-2}\right\}\\ =\frac{1}{(x-2)}\left\{\frac{x-1}{x-2}\right\}\end{array}$

  $\begin{array}{l}=\frac{2x-4+2}{x(x-2)}\\ =\frac{2x-2}{x(x-2)}\end{array}$

New term $\frac{2x-2}{x(x-2)}$ (i)

Third term

  $=\frac{2}{x^2-x-2}$

Denominator = $x^2-x-2$

where $r=-2$ and $s=1$ .

  $\begin{array}{l}r+s=-2+1=-1\\ rs=(-2)(1)=-2\\ (x+r)(x+s)=(x-2)(x+1)\end{array}$  (ii)

Solving equation (i) and (ii)

  $=\frac{2x-2}{x(x-2)}+\frac{2}{(x-2)(x+1)}$

Taking $\frac{1}{(x-2)}$ common

  $=\frac{1}{(x-2)}\left\{\frac{2x-2}{x}+\frac{2}{(x+1)}\right\}$

writing the fraction in LCD (least common denominator) form we have,

  $\begin{array}{l}=\frac{1}{(x-2)}\left\{\frac{(2x-2)(x+1)}{x(x+1)}+\frac{2x}{x(x+1)}\right\}\\ =\frac{1}{(x-2)}\left\{\frac{2x^2+2x-2x-2}{x(x+1)}+\frac{2x}{x(x+1)}\right\}\\ =\frac{1}{(x-2)}\left\{\frac{2x^2-2+2x}{x(x+1)}\right\}\\ =\frac{1}{(x-2)}\left\{\frac{2x^2+2x-2}{x(x-2)}\right\}\end{array}$

  $=\frac{2x^2+2x-2}{x{(x-2)}^2}$

After simplifying we get $\frac{2x^2+2x-2}{x{(x-2)}^2}$ .