Question

\frac{x-5}{x^4-2x^3}\cdot \frac{x^2-4}{x^2-3x-10} x 4 −2x 3 x−5 ​ ⋅ x 2 −3x−10 x 2 −4 ​

Answer 1

Answer:

$\frac{x-5}{x^4-2x^3}\cdot \frac{x^2-4}{x^2-3x-10} = \frac{1}{x^3}$

Step-by-step explanation:

Given

$\frac{x-5}{x^4-2x^3}\cdot \frac{x^2-4}{x^2-3x-10}$

Required

Solve

Express $x^2-4$ as difference of two squares

$\frac{x-5}{x^4-2x^3}\cdot \frac{(x-2)(x+2)}{x^2-3x-10}$

Factorize $x^4 - 2x^3$

$\frac{x-5}{x^3(x-2)}\cdot \frac{(x-2)(x+2)}{x^2-3x-10}$

Cancel out x - 2

$\frac{x-5}{x^3}\cdot \frac{x+2}{x^2-3x-10}$

Expand $x^2 - 3x - 10$

$\frac{x-5}{x^3}\cdot \frac{x+2}{x^2+2x-5x-10}$

Factorize:

$\frac{x-5}{x^3}\cdot \frac{x+2}{x(x+2)-5(x+2)}$

Factor out x + 2

$\frac{x-5}{x^3}\cdot \frac{x+2}{(x-5)(x+2)}$

Cancel out x - 5 and x + 2

$\frac{1}{x^3}\cdot \frac{1}{1}$

$\frac{1}{x^3}\cdot 1$

$\frac{1}{x^3}$

Hence:

$\frac{x-5}{x^4-2x^3}\cdot \frac{x^2-4}{x^2-3x-10} = \frac{1}{x^3}$