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

1 Answer
Mar 11, 2017

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#