Question

Give the factors for the numerator [n] and denominator [d] after reducing.

Answer 1

Answer:

The factors for the numerator [n] and denominator [d] after reducing will be:

$\frac{x^2-25}{x^2-4x}\div \:\frac{2x^2+2x-40}{x^3-x}=\frac{\left(x-5\right)\left(x+1\right)\left(x-1\right)}{2\left(x-4\right)^2}$

Step-by-step explanation:

Given the expression

$\frac{x^2-25}{x^2-4x}\div \frac{2x^2+2x-40}{x^3-x}$

$\mathrm{Apply\:the\:fraction\:rule}:\quad \frac{a}{b}\div \frac{c}{d}=\frac{a}{b}\times \frac{d}{c}$

$=\frac{x^2-25}{x^2-4x}\times \frac{x^3-x}{2x^2+2x-40}$

$\mathrm{Multiply\:fractions}:\quad \frac{a}{b}\times \frac{c}{d}=\frac{a\:\times \:c}{b\:\times \:d}$

$=\frac{\left(x^2-25\right)\left(x^3-x\right)}{\left(x^2-4x\right)\left(2x^2+2x-40\right)}$

$=\frac{\left(x^2-25\right)x\left(x^2-1\right)}{\left(x^2-4x\right)\left(2x^2+2x-40\right)}$

$\mathrm{Cancel\:the\:common\:factor:}\:x$

$=\frac{\left(x^2-25\right)\left(x^2-1\right)}{2\left(x-4\right)\left(x^2+x-20\right)}$

∵ As factor  $\left(x^2-25\right)\left(x^2-1\right)=\left(x+5\right)\left(x-5\right)\left(x+1\right)\left(x-1\right)$

so the expression becomes

$=\frac{\left(x+5\right)\left(x-5\right)\left(x+1\right)\left(x-1\right)}{2\left(x-4\right)\left(x^2+x-20\right)}$

∵ As factor $2\left(x-4\right)\left(x^2+x-20\right)=2\left(x-4\right)^2\left(x+5\right)$

so the expression becomes

$=\frac{\left(x+5\right)\left(x-5\right)\left(x+1\right)\left(x-1\right)}{2\left(x-4\right)^2\left(x+5\right)}$

$\mathrm{Cancel\:the\:common\:factor:}\:x+5$

$=\frac{\left(x-5\right)\left(x+1\right)\left(x-1\right)}{2\left(x-4\right)^2}$

Therefore, the factors for the numerator [n] and denominator [d] after reducing will be:

$\frac{x^2-25}{x^2-4x}\div \:\frac{2x^2+2x-40}{x^3-x}=\frac{\left(x-5\right)\left(x+1\right)\left(x-1\right)}{2\left(x-4\right)^2}$