Question

Simplify the following polynomials in the simplest form:

$\frac{1}{x^2+3x+2}+\frac{1}{x^2+5x+6}-\frac{2}{x^2+4x+3}$

Answer 1

The given expression $\frac{1}{x^2+3x+2}+\frac{1}{x^2+5x+6}-\frac{2}{x^2+4x+3}$can be simplified as shown below:

$=\frac{1}{x^2+3x+2}+\frac{1}{x^2+5x+6}-\frac{2}{x^2+4x+3}$

$=[\frac{1}{x^2+3x+2}+\frac{1}{x^2+5x+6}]-\frac{2}{x^2+4x+3}$

$=[\frac{1}{x^2+x+2x+2}+\frac{1}{x^2+3x+2x+6}]-\frac{2}{x^2+x+3x+3}$

$=[\frac{1}{x(x+1)+2(x+1)}+\frac{1}{x(x+3)+2(x+3)}]-\frac{2}{x(x+1)+3(x+1)}$

$=[\frac{1}{(x+2)(x+1)}+\frac{1}{(x+2)(x+3)}]-\frac{2}{(x+3)(x+1)}$

$=[\frac{(x+3)}{(x+1)(x+2)(x+3)}+\frac{(x+1)}{(x+1)(x+2)(x+3)}]-\frac{2}{(x+1)(x+2)(x+3)}\left(TakingLCM\right)$

$=\left[\frac{2x+4}{(x+1)(x+2)(x+3)}\right]-\frac{2}{(x+1)(x+2)(x+3)}$

$=\left[\frac{2(x+2)}{(x+1)(x+2)(x+3)}\right]-\frac{2}{(x+1)(x+2)(x+3)}$

$=\frac{2}{(x+1)(x+2)(x+3)}-\frac{2}{(x+1)(x+2)(x+3)}$

$=0$

Hence, $\frac{1}{x^2+3x+2}+\frac{1}{x^2+5x+6}-\frac{2}{x^2+4x+3}=0$.