Question

How to factor this?
4x^4n+2 -28x^2n+1y^3n + 49y^6n

Answer 1

Take the gcf first

Answer 2

There isn't one... just 1

Answer 3

Everything is multiplied by n

Answer 4

except 2

Answer 5

\(4x^4n+2 -28x^2n+1y^3n + 49y^6n\) or are the ns part of the exponents?

Answer 6

Right but..how would you pull that out?

Answer 7

n is also a part of the exponents @e.mccormick

Answer 8

Oh, that makes it a bit more complicated..

Answer 9

You see, I was taught that you solve the same way just replace with you know the whole x y whatever like a normal trinomial. Then at the end you divide the exponents by two..I don't get it lol

Answer 10

\(4x^{4n}+2 -28x^{2n}+1y^{3n} + 49y^{6n}\) ??

Answer 11

that +2 is a part of the exponent as well

Answer 12

Let me guess. The +2 is in an exponent too.

Answer 13

4x^{4n+2} -28x^{2n}+1y^{3n} + 49y^{6n}

Answer 14

yes.

Answer 15

\(4x^{4n+2} -28x^{2n}+1y^{3n} + 49y^{6n}\)

Answer 16

wait

Answer 17

\[4x^{4n+2}-28^{2n}+y^{3n} + 49y^{6n}\]

Answer 18

that?

Answer 19

that y^3n is attached to -28x^3n

Answer 20

but that y is not the exponent.

Answer 21

oh.
\[4x^{4n+2}-28^{2n}y^{3n}+49y^{6n}\]
this?

Answer 22

yesss

Answer 23

Alright, let's take the gcf of the factors after the 4x^(4n+2) then...

\[4x^{4n+2}-28^{2n}y^{3n}+49y^{6n}\]
\[4x^{4n+2}7y^{3n}(-4^{2n}+7^{3n})\]

not sure if taking the ^3n is allowed but this is what i got.

Answer 24

how..

Answer 25

Did you forget a - there?

Answer 26

the -1?
you could take it out making (4n^2n - 7^3n) i think ... .-.

Answer 27

Uh..

Answer 28

\(4x^{4n+2}-7y^{3n}(4^{2n}-7y^{3n})\) was what I meant.

Answer 29

yeah you could do that.

Answer 30

But then are you factoring from there?

Answer 31

I think we lost part of the original.

Answer 32

mick , you attached the y^3n .. to the 7, i thought it was factored out.

Answer 33

\(4x^{4n+2} -28x^{2n+1}y^{3n} + 49y^{6n}\)
Now it will make more sense.

Answer 34

oh.

Answer 35

I'm completely lost..

Answer 36

When we "fixed" the way it looked, we had lost a +1 in on of the exponents. That is critical.

\(4x^{4n+2} -28x^{2n+1}y^{3n} + 49y^{6n} \implies \)

\(4x^{2(2n+1)} -7 y^{3n}(4x^{2n+1} + 7y^{3n}\)

Answer 37

Yeah. dunno how you write in latex so fast >.> it takes me forever.

Answer 38

@e.mccormick is some kind of \(\LaTeX\) master. That's why he types so faset.

Answer 39

Bah, forgot the sign change! LOL

\(4x^{4n+2} -28x^{2n+1}y^{3n} + 49y^{6n}\implies \)

\(4x^{2(2n+1)} -7 y^{3n}(4x^{2n+1} - 7y^{3n}) \implies \)

\(4x^{2n+1}x^2 -7 y^{3n}(4x^{2n+1} - 7y^{3n}) \implies \)

\((x^2 -1)(4x^{2n+1} + 7y^{3n}) \)

Answer 40

Cause I type it rather than use the editor.

Answer 41

Was about to write that, but the Latex God , did it before me haha lol.

Answer 42

\((x^2 - 1)\) can be factored further

Answer 43

Yeah,
(x+1)(x-1)

Answer 44

Yep. Now, KC, you get what was done there? Too much confusion with what it was at the start... but now I thing it can be explained.

Answer 45

I uh..uhm. er.

Answer 46

Would you like to see how I solved it?

Answer 47

OK

Answer 48

Er well actually I have to go somewhere, so I can't. Its alright I will get help from my dad later. Thank you all for helping!

Answer 49

Hehe. Kk.

Answer 50

Aaargh! I was looking at it, and something seemed wrong... it is a perfect square trinomial!

Answer 51

Yep, that's what I saw as well, but I didn't say anything because I figured you knew what you're doing so who am I to argue.

Answer 52

Lol I figured out how to do it on my own no problem :)