Question

what is (1-(4/x))/(1-(2/x)-(8/x^2)) in simplest form? please explain how you get your answer

Answer 1

also known as:\[\frac{ 1-\frac{ 4 }{ x } }{ 1-\frac{ 2 }{ x }-\frac{ 8 }{ x ^{2} } }\]

Answer 2

\[\huge\frac{1-\frac{4}{x}}{1-\frac{2}{x}-\frac{8}{x^2}}\]

Answer 3

same idea as before find a common denominator

Answer 4

get the answer is one step by multiplying top and bottom by \(x^2\) cancelling merrily as you go

Answer 5

ok

Answer 6

what satellite73 said works too, and faster

Answer 7

so...\[\frac{ \frac{ x ^{2} }{ x ^{2} }-\frac{ 4x }{ x ^{2} } }{ \frac{ x ^{2} }{ x ^{2} }-\frac{ 2x }{ x ^{2} }-\frac{ 8 }{ x ^{2} } }\]?

Answer 8

oh no !!

Answer 9

what?

Answer 10

\[\huge\frac{(1-\frac{4}{x})}{(1-\frac{2}{x}-\frac{8}{x^2})}\times \frac{x^2}{x^2}\]

Answer 11

distribute
what is \(1\times x^2\) ?

Answer 12

\[x ^{2}\]

Answer 13

ok and \(-\frac{4}{x}\times x^2\) ?

Answer 14

i dont know

Answer 15

really? don't believe you
\[\frac{-4x^2}{x}=?\]?

Answer 16

im confused im sorry

Answer 17

cancel an x top and bottom or subtract the exponent
either way you get
\[-\frac{4}{x}\times x^2=-4x\]

Answer 18

oh ok

Answer 19

so the numerator is
\[x^2-4x\]
how about the denominator?

Answer 20

ok \[x ^{2}-2x-8x\]is the denominator?

Answer 21

close

Answer 22

\[x^2-2x-8\] the last term is just \(-8\) cause the \(x^2\) cancel

Answer 23

oh right

Answer 24

so we are at
\[\frac{x^2-4x}{x^2-2x-8}\] now i think we can factor top and bottom

Answer 25

ok

Answer 26

matter of fact it am sure of it
do you see how to factor the top?

Answer 27

yes it would be x(x-4)

Answer 28

and the bottom?

Answer 29

and the bottom would be (x-4)(x+2)

Answer 30

then cancel and be done!

Answer 31

then you can cancel out the (x-4)s
ok thank you

Answer 32

yw