Answer:
[tex]\frac{x-5}{x^4-2x^3}\cdot \frac{x^2-4}{x^2-3x-10} = \frac{1}{x^3}[/tex]
Step-by-step explanation:
Given
[tex]\frac{x-5}{x^4-2x^3}\cdot \frac{x^2-4}{x^2-3x-10}[/tex]
Required
Solve
Express [tex]x^2-4[/tex] as difference of two squares
[tex]\frac{x-5}{x^4-2x^3}\cdot \frac{(x-2)(x+2)}{x^2-3x-10}[/tex]
Factorize [tex]x^4 - 2x^3[/tex]
[tex]\frac{x-5}{x^3(x-2)}\cdot \frac{(x-2)(x+2)}{x^2-3x-10}[/tex]
Cancel out x - 2
[tex]\frac{x-5}{x^3}\cdot \frac{x+2}{x^2-3x-10}[/tex]
Expand [tex]x^2 - 3x - 10[/tex]
[tex]\frac{x-5}{x^3}\cdot \frac{x+2}{x^2+2x-5x-10}[/tex]
Factorize:
[tex]\frac{x-5}{x^3}\cdot \frac{x+2}{x(x+2)-5(x+2)}[/tex]
Factor out x + 2
[tex]\frac{x-5}{x^3}\cdot \frac{x+2}{(x-5)(x+2)}[/tex]
Cancel out x - 5 and x + 2
[tex]\frac{1}{x^3}\cdot \frac{1}{1}[/tex]
[tex]\frac{1}{x^3}\cdot 1[/tex]
[tex]\frac{1}{x^3}[/tex]
Hence:
[tex]\frac{x-5}{x^4-2x^3}\cdot \frac{x^2-4}{x^2-3x-10} = \frac{1}{x^3}[/tex]
