Question

Can someone help me out on multiplying and simplifying x-6/x^2 + 4x - 32 * x-4/x+2 ?

Answer 1

\[\large \frac{x-6}{x^2+4x-32}\times \frac{x-4}{x+2}\]?

Answer 2

yes!

Answer 3

not much to do
maybe we can factor the denominator and cancel something

Answer 4

yeah i think that's what it calls for, but i'm so confused on how to do any of that

Answer 5

\[x^2+4x-32=(x+8) (x-4)\]

Answer 6

?

Answer 7

so you do get to cancel one factor but that is it
\[\large \frac{x-6}{x^2+4x-32}\times \frac{x-4}{x+2}\\
=\frac{(x-6)(x-4)}{(x+8)(x-4)(x+2)}\\
=\frac{(x-6)}{(x+8)(x+2)}\]

Answer 8

cancelled the common factor top and bottom of \(x-4\)

Answer 9

OHHHH, this i understand! :D the notes i have on it are so confusing. thank you so much, can you help me with two more? or explain the rule for it?

Answer 10

sure
you want to post it here?

Answer 11

x^2 - 4/x^2 - 1 * x+1/x+2

Answer 12

\[\frac{x^2-4}{x^2-1}\times \frac{x+1}{x+2}\] again factor and cancel
\[\large \frac{(x+2)(x-2)(x+1)}{(x+1)(x-1)(x+2)}\]

Answer 13

this time you can cancel two factors
clear?

Answer 14

yeah i think so

Answer 15

both factoring was from the difference of two square
\[a^2-b^2=(a+b)(a-b)\] so
\[x^2-4=(x+2)(x-2)\] in the top and
\[x^2-1=(x+1)(x-1)\] in the bottom

Answer 16

you are left with \[\large \frac{\cancel{(x+2)}(x-2)\cancel{(x+1)}}{\cancel{(x+1)}(x-1)\cancel{(x+2)}}\]\[=\frac{x-2}{x-1}\]

Answer 17

I think I'm getting it, wow I'm surprised! Can you help me with just this last one then I'll stop bugging you?

Answer 18

k no problem

Answer 19

x+3/x * x+4/x^2 + 7x + 12

Answer 20

\[\frac{x+3}{x}\times \frac{x+4}{x^2+7x+12}\]?

Answer 21

yeah, this one is a bit confusing :/

Answer 22

this one you can cancel your brains out because
\[x^2+7x+12=(x+3)(x+4)\]

Answer 23

\[\frac{x+3}{x}\times \frac{x+4}{x^2+7x+12}\]\[=\frac{x+3}{x}\times \frac{x+4}{(x+3)(x+4)}\]

Answer 24

now everything goes
leaves only \(1\) in the top and \(x\) in the bottom

Answer 25

\[=\frac{(x+3)(x+4)}{x(x+3)(x+4)}\]
\[=\frac{1}{x}\]

Answer 26

I think I am proud to say I fully understand it now! It took me forever by just looking at my textbook. Thanks for the help :)

Answer 27

yw
glad to help
good luck!