Question

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

Answer 1

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

`(x^2-25)/(x^2-4x)/ \:(2x^2+2x-40)/(x^3-x)=(\left(x-5\right)\left(x+1\right)\left(x-1\right))/(2\left(x-4\right)^2)`

Explanation:

Given the expression

`(x^2-25)/(x^2-4x)/ (2x^2+2x-40)/(x^3-x)`

`\mathrm{Apply\:the\:fraction\:rule}:\quad (a)/(b)/ (c)/(d)=(a)/(b)* (d)/(c)`

`=(x^2-25)/(x^2-4x)* (x^3-x)/(2x^2+2x-40)`

`\mathrm{Multiply\:fractions}:\quad (a)/(b)* (c)/(d)=(a\:* \:c)/(b\:* \:d)`

`=(\left(x^2-25\right)\left(x^3-x\right))/(\left(x^2-4x\right)\left(2x^2+2x-40\right))`

`=(\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`

`=(\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

`=(\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

`=(\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`

`=(\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:

`(x^2-25)/(x^2-4x)/ \:(2x^2+2x-40)/(x^3-x)=(\left(x-5\right)\left(x+1\right)\left(x-1\right))/(2\left(x-4\right)^2)`