Question

Can anyone help walk thru simplifying expressions?

Answer 1

(x^2- 4x - 21) / (x^2- 6x + 8) * (x - 4) / (x^2 - 2x - 35)

Answer 2

\[\large \frac{x^{2} - 4x - 21}{x^{2} - 6x + 8} \mathsf{x} \frac{x-4}{x^{2} - 2x - 35}\]

this is the problem right?

Answer 3

Yes

Answer 4

I didn't know how to post it in the way you did :o)

Answer 5

hahaha..it takes time to learn :P what you did is just fine..anyway..do you know how to factor out each one of them?

Answer 6

Well....I'm not sure of the right way as our teacher is a bit odd. She had us using the Hong Kong Method yesterday.....does that make sense?

Answer 7

i do not know what the hongkong method is...but let's just factor each one individually...i'll do one...

x^2 - 4x - 21 ---> x^2 + 3x - 7x - 21 (i just split the middle term)

x(x+3) - 7(x+3) -> (x+3)(x-7)

now...can you do the denominator?

Answer 8

OK Hold a sec lemme see what I have too

Answer 9

for the top left I have (x+3)(x-7)

Answer 10

for the lower left I have (x-2)(x-4)

Answer 11

for the lower right is (x+5)(x-7)

Answer 12

I did nothing to the upper right, as not sure about it

Answer 13

upper right is already factored so....lemme write all our info..

Answer 14

ok

Answer 15

\[\large \frac {(x+3)(x-7)}{(x-2)(x-4)} \mathsf{x} \frac {x-4}{(x+5)(x-7)}\]

look how you can cancel the x -4 on the bottom left and the top right...

cancel the x - 7 on the upper left and lower right too...so you're left with...

\[\large \frac {x+3}{x-2} \mathsf{x} \frac{1}{x+5}\]

if we simplify that..we have...

\[\huge \frac {(x+3)(1)}{(x-2)(x+5)}\]

right?

Answer 16

Thanks SO very much...I will try a few more & if I get stuck is it ok to reply to you again?

Answer 17

sure :)

Answer 18

Great!

Answer 19

Is there a way on this site for me to LIKE your avatar/nickname etc so I can know who has been helping me? I am Stacy, btw.....trying to help my son & he's watching too :)

Answer 20

uhmmm you can fan someone...which you just did...lol

Answer 21

LOL...just found it...thanks

Answer 22

Well......now he's stuck on a^2-b^2/3a^2-6a+3*a^2-1/4a+4b

Answer 23

hmmm...

\[\large \frac{a^{2} - b^{2}}{3a^{2} - 6a + 3} \mathsf{x} \frac{a^{2} - 1}{4a + 4b}\]

like in the first problem...we factor out each one...

i'll do the first one...

a^2 - b^2 = (a+b)(a-b)

can you do the next ones?

Answer 24

he has no idea how you got that

Answer 25

I was never good in math/alg either :(

Answer 26

which one? a^2 - b^2 = (a+b)(a-b)? it is a property...difference of two squares....that is the property itself..a^2 -b^2 = (a+b)(a-b)

Answer 27

no idea :(

Answer 28

if you'd like it broken down (though i think it's worth that he knows that shortcut)...look at this.. (a+b)(a-b) by distibutive property...(a)(a) + (ab) - (ab) + (b)(b)

= a^2 + ab - ab + b^2

ab gets canceled out...

a^2 + b^2

Answer 29

he's not even sure what is :/ this teacher is odd to say the least....never teaches the same thing 2 days in a row :( I don't get it either

Answer 30

we r assuming (a)(a) is a^2 and the same for (b)(b)...but not sure on the (ab) (ab)

Answer 31

(a+b)(a-b)..FOIL method..

multiply the First terms -> (a)(a)
multiply the Outside terms -> (a)(-b)
multiply the Inside terms -> (a)(b)
multiply the Last terms -> (b)(-b)

add those up...

(a)(a) + (a)(-b) + (a)(b) + (b)(-b)
a^2 - ab + ab - b^2
a^2 - b^2 + (ab- ab)....according to the addiitive inverse...anything added to it's negative equals zero (i.e. 1-1 = 0; 2-2 = 0; 3-3 = 0)

so...

a^2 - b^2 + 0
a^2 - b^2

making sense now?

Answer 32

he says kinda

Answer 33

ahh good. so can you factor out the remaining ones now?

Answer 34

:( dun think so

Answer 35

he's completely giving up, but thanks for all of your help this evening. My son has Crohns Disease and is easily frustrated...

Answer 36

his meds don't help make it any better

Answer 37

ohhh...well tell him you're welcome and i tried my best :)