How do you divide #\frac { 1- \frac { 4} { x } } { 1- \frac { 2} { x } - \frac { 8} { x ^ { 2} } }#?

2 Answers
Dec 7, 2016

#x/(x + 2)#

Explanation:

First, you must get the terms in the numerator over a common denominator and the terms in the denominator over a common denominator by multiplying each term by the correct for of #1#:

#((x/x) * 1 - 4/x)/((x^2/x^2) * 1 - (x/x)*(2/x) - 8/x^2)#

#(x/x - 4/x)/(x^2/x^2 - (2x)/x^2 - 8/x^2)#

#((x - 4)/x)/((x^2 - 2x - 8)/x^2)#

Now we can factor the numerator of the bottom fraction:

#((x - 4)/x)/(((x+ 2)(x - 4))/x^2)#

We can now us the rule for dividing fractions to divide these fractions. Remember: #color(red)((a/b)/(c/d) = (a*d)/(b*c))#

#((x - 4)x^2)/(x(x + 2)(x - 4))#

#(cancel((x - 4))x^2)/(x(x + 2)cancel((x - 4)))#

#(x^2)/(x(x + 2))#

#x^(2-1)/(x + 2)#

#x/(x + 2)#

Dec 7, 2016

#x/(x+2)#

Explanation:

Write as:#" " [1-4/x] -:[1-2/x-8/x^2]#

#" "[(x-4)/x]-:[(x^2-2x-8)/x^2]#

#" "((x-4))/x xx x^2/((x+2)(x-4))#

#" "(cancel((x-4)))/(cancel(x)) xx x^(cancel(2))/((x+2)cancel((x-4)))#

#color(white)(.)#

#" "x/(x+2)#