Question

Can someone please help me with this?
dividing rational expressions

Answer 1

sure...

Answer 2

Can you please walk me through this?

Answer 3

We need to work on one thing at a time :)

Answer 4

So, let's only focus on the numerator part of the big fraction (:

Answer 5

OK:)

Answer 6

\(\sf\Large \frac{x^2-3x-88}{x^2-49} =?\)

The denominator of this looks easy to factor, considering the rule:
\(\sf \Large a^2 - b^2 = (a+b)(a-b)\)

Answer 7

Hint for factoring the numerator:

8 + (-11) = -3
8 x (-11) = -88

Answer 8

Ok, here is what I got factored:
\[\frac{ (x-11)(x+8) }{ (x-7)(x+7) }\]

Answer 9

Good job! :)

Answer 10

:)

Answer 11

Now, let's factor the denominator.

\(\bf\Huge \frac{x-11}{x^2-2x-63}\)

Hint: the numerator is already factored
the denominator's hints are:

7 + (-9) = -2
7 x (-9) = -63

Answer 12

Now I got:
\[\frac{ x-11 }{ (x-9)(x+7) }\]

Answer 13

Correct, so now we have:

\(\bf\Large\frac{ (x-11)(x+8) }{ (x-7)(x+7)}\div \frac{ x-11 }{ (x-9)(x+7) }\)

Answer 14

And we can simplify that too:

\(\bf\Large\frac{ (x-11)(x+8) }{ (x-7)(x+7)}\times \frac{ (x-9)(x+7) }{ x-11 }\)

Answer 15

Simplify it more:

\(\bf\Large\frac{\cancel{ (x-11)}(x+8) }{ (x-7)\cancel{(x+7)}}\times \frac{ (x-9)\cancel{(x+7)} }{ \cancel{(x-11)} }\)

Answer 16

Oh, so that would make:
\[\frac{ (x+8)(x-9) }{ (x-7) }\]

Answer 17

And that's your answer :D

Answer 18

:D Thanks!

Answer 19

Anytime! :D