Question

Can someone explain to me how 6x^2(x + 4)^5 / 9x^3(x + 4) equals to 2(x + 4)^4 / 3x

Answer 1

\[\huge\rm \frac{ 6x^2(x+4)^5 }{ 9x^3(x+4) } = \frac{ 2(x+4)^4 }{ 3x }\]
you can cancel out few things here hmm

Answer 2

All I know is that 6 and 9 is 2/3 and x3 subtracts with x2, but i dont get how you get the 4th power

Answer 3

and how the x+4 dont cancel out

Answer 4

\[\huge\rm \frac{ \color{red}{6x^2} \color{green}{(x+4)} \times (x+4)^4 }{ \color{red}{9x^3}\color{green}{(x+4)} } = \frac{ 2(x+4)^4 }{ 3x }\]
you can cancel out few things here hmm

Answer 5

when you multiply same variables number you should add their exponent \[\huge\rm x^m + x^n = x^{m+n}\]

Answer 6

\(\color{blue}{\text{Originally Posted by}}\) @MUSTARD_MK
All I know is that 6 and 9 is 2/3 and x3 subtracts with x2, but i dont get how you get the 4th power
\(\color{blue}{\text{End of Quote}}\)
which one ?? at the right side ?

Answer 7

Yes sorry

Answer 8

nah that's fine
it's in the question

Answer 9

\(\color{blue}{\text{Originally Posted by}}\) @MUSTARD_MK
Can someone explain to me how 6x^2(x + 4)^5 / 9x^3(x + 4) equals to 2(x + 4)^4 / 3x
\(\color{blue}{\text{End of Quote}}\)
here
equals to 2(x+4)^4 <----

Answer 10

yup

Answer 11

i dont get how it went from a 5 to a 4, that really confuses me

Answer 12

I feel dumb lol

Answer 13

ohh left side ?
like i said when you multiply same variables you should add their exponent
like \[\huge\rm \frac{ x^2 }{ x^3 } = \frac{ \cancel x \times \cancel x }{ x \times \cancel x \times \cancel x }\]

Answer 14

so that's how (x+4)^5 can be written as (x+4) (x+4)^4
let's try by this way
applying exponent rule \[\huge\rm \frac{ x^m }{ x^n } = x^{m -n}\]
when bases are same you can move exponent from bottom to the top
so \[\large\rm \frac{ (x+4)^5 }{ (x+4)^1 } = (x+4)^{5-1}\]

Answer 15

Oooooooooh

Answer 16

So theres like a hidden 1 there?

Answer 17

sort of

Answer 18

yes thief 1 :/
:-)
yes one is always there
so
if i ask what's the exponent of (x) answer would be ?

Answer 19

1

Answer 20

- (x+ 5)
how would you solve this distribute by what ?
example

Answer 21

-x -5 right?

Answer 22

yes so distribute by -1 right ?
so yes 1 is always there :-)

Answer 23

awesome

Answer 24

back to the question \[\huge\rm \frac{ \color{red}{6x^2} \color{green}{(x+4)} \times (x+4)^4 }{ \color{red}{9x^3}\color{green}{(x+4)} } = \frac{ 2(x+4)^4 }{ 3x }\]

Answer 25

I don't want to piss you off, but where did the extra (x+4) from the top left came from?

Answer 26

hahha it's okay gO_Od to ask a question :-)
i like that
i just separated that

Answer 27

Oooooh

Answer 28

thats smart lol

Answer 29

okay we can write like this \[\huge\rm \frac{ \color{red}{6x^2} \times (x+4)^5 }{ \color{red}{9x^3}\color{green}{(x+4)^1} } = \frac{ 2(x+4)^4 }{ 3x }\]
apply exponent rule

Answer 30

ah, that makes a bit more sense now for me

Answer 31

yeah i thought you are familiar with exponent rules so it would help you to understand :-)

Answer 32

Thanks man, I really appreciate it :)

Answer 33

got it ?? :o

Answer 34

finish ?? :o

Answer 35

Yeah, I understand how to do it now

Answer 36

ohh okay nice ! gO_Od luck!!!

Answer 37

Gracias!

Answer 38

if you want me to check ur answer just tag meh :-)

Answer 39

I'll keep that noted

Answer 40

no hay problema