How do I divide this rational expression?

Question

9x^3/x^3-x^2 divided by x-8/x^2-9x+8

Answer 1

a/b ÷ c/d = a/b · d/c = a·d / b·c, so

9x²/x³-x² divided by x-8/x²-9x+8 =

9x³·(x²-9x+8) / (x³-x²)·(x-8) =

{now factor top and bottom as far as possible}

9x³·(x-8)(x-1) / x²·(x-1)·(x-8) =

{now (x-8)(x-1) on the top cancels with (x-8)(x-1) on the bottom. Also x³/ x² simplifies to x }

9x

Hope this helps.

Answer 2

You have to simplify your complex fraction before division is possible.

x^3 - x^2 becomes x^2(x - 1)

We now have this guy:

9x^3/(x^2)(x - 1)

Next:

Factor the trinomial denominator. See it?

x^2 - 9x + 8 becomes (x - 1) (x - 8)

We now have this guy:

1/(x-1)

NOTE: The quantity (x - 8) cancels itself out (division rule).

We now have this complex fraction:

9x^3/(x^2)(x - 1) divided by 1/(x-1)

Just like regular division, we INVERT the right side fraction and it becomes (x-1)/1 and then we multiply it by the other fraction (the one on the left side).

Doing so, we get this:

9x^3/x^2

We now divide x^3 by x^2 and we get x (on the numerator).

Final answer:

9x

Guido

Answer 3

9x^3/(x^3-x^2)
= 9x^3/[x^2(x-1)]
= 9x / (x-1)

(x-8) / (x^2-9x+8)
=(x-8) / [(x-8)(x-1)]
=1/(x-1)

[9x^3/(x^3-x^2)] / [(x-8) / (x^2-9x+8)]
= [9x / (x-1)] / [1/(x-1)]
= [9x / (x-1)] * (x-1)
= 9x

Answer 4

first you need to factor:
[(9x^3)/(x^3-x^2)] / [(x-8)/(x^2-9x+8)] =
[(9x^3)/((x^2)(x-1))] / [(x-8)/((x-1)((x-8)] =
then simplify:
[(9x)/(x-1)] / [1/(x-1)] =
[(9x)/(x-1)] *(x-1) =
9x(x-1)/(x-1) =
9x

Answer 5

9x^3 / x^2(x - 1) = 9x / x - 1

x-8/(x-8)(x-1) = 1/x-1

Answer 6

9x^3 ............. x - 8
------------ ÷ ------------------ =
x²(x - 1) ...... (x - 8)(x - 1)

9x^3 ............. x - 1
------------ x ------------------ =
x²(x - 1) ............ 1

9x^3
------- = 9x
x²