#### Question

Solve the following quadratic equations by factorization:\[\frac{1}{x - 3} + \frac{2}{x - 2} = \frac{8}{x}; x \neq 0, 2, 3\]

#### Solution

\[\frac{1}{x - 3} + \frac{2}{x - 2} = \frac{8}{x}\]

\[ \Rightarrow \frac{\left( x - 2 \right) + 2\left( x - 3 \right)}{\left( x - 3 \right)\left( x - 2 \right)} = \frac{8}{x}\]

\[ \Rightarrow \frac{x - 2 + 2x - 6}{x^2 - 2x - 3x + 6} = \frac{8}{x}\]

\[ \Rightarrow \frac{3x - 8}{x^2 - 5x + 6} = \frac{8}{x}\]

\[ \Rightarrow x\left( 3x - 8 \right) = 8\left( x^2 - 5x + 6 \right)\]

\[ \Rightarrow 3 x^2 - 8x = 8 x^2 - 40x + 48\]

\[ \Rightarrow 5 x^2 - 32x + 48 = 0\]

\[ \Rightarrow 5 x^2 - 20x - 12x + 48 = 0\]

\[ \Rightarrow 5x\left( x - 4 \right) - 12\left( x - 4 \right) = 0\]

\[ \Rightarrow \left( 5x - 12 \right)\left( x - 4 \right) = 0\]

\[ \Rightarrow 5x - 12 = 0 \text { or } x - 4 = 0\]

\[ \Rightarrow x = \frac{12}{5} or x = 4\]

Hence, the factors are 4 and \[\frac{12}{5}\].