Question

How do you divide `\frac { ( x - 5) ( x + 5) } { 4x ^ { 4} } \div \frac { 2x - 10} { 8x ^ { 5} }`?

Answer 1

See the entire solution process below:

Explanation:

First, rewrite the term on the right as:

`((x - 5)(x + 5))/(4x^4) -: (2(x - 5))/(8x^5)`

Then rewrite the expression as:

`(((x - 5)(x + 5))/(4x^4))/((2(x - 5))/(8x^5))`

We can now rewrite this expression again using this rule for dividing fractions:

`(color(red)(a)/color(blue)(b))/(color(green)(c)/color(purple)(d)) = (color(red)(a) xx color(purple)(d))/(color(blue)(b) xx color(green)(c))`

`(color(red)((x - 5)(x + 5))/color(blue)((4x^4)))/(color(green)(2(x - 5))/color(purple)((8x^5))) = (color(red)((x - 5)(x + 5)) xx color(purple)((8x^5)))/(color(blue)((4x^4)) xx color(green)(2(x - 5)))`

We can next cancel some of the common terms in the numerator and denominator:

`(color(red)(cancel((x - 5))(x + 5)) xx color(purple)((cancel(8)x^5)))/(color(blue)((cancel(4)x^4)) xx color(green)(cancel(2)cancel((x - 5)))) = (x^5(x + 5))/x^4`

We can now use these rules of exponents to further simplify the result:

`x^color(red)(a)/x^color(blue)(b) = x^(color(red)(a)-color(blue)(b))` and `a^color(red)(1) = a`

`(x^color(red)(5)(x + 5))/x^color(blue)(4) = x^(color(red)(5)-color(blue)(4))(x + 5) = x^1(x + 5) =`

`x(x + 5)`

Or

`x^2 + 5x`

However, `(2(x - 5))/(8x^5) != 0`