Question

how do you factor a^2^n +6a^n+8

Answer 1

Just to be clear, is it really a\(\sf\color{red}{^{2n}}\)+a\(\sf\color{red}{^{n}}\)+8?

Answer 2

yes

Answer 3

6a^n sorry

Answer 4

Rewrite it using LaTex (the equation editor) on the bottom of this little box.

Answer 5

\[a ^{2n}+6a ^{n}+8\]

Answer 6

Well, are you solving for \(a\) or for \(n\)?

Answer 7

It says to factor completely or indicate if prime

Answer 8

Well, eitheway, what two numbers when multiplied give you 8, but when added equal 6?

Answer 9

I am going to assume that n = some number (for simplicity, i will assume n = 1)

Answer 10

ok

Answer 11

Are you with me so far? B/c 6 or 8 is nOT the answer.

Answer 12

yes

Answer 13

Well, what are those two numbers @red32968

___ \(\times\) ___ = 8
___ + ___ = 6

Answer 14

4*2
4+2

Answer 15

Yes, Correct! So you have: a\(^{2n}\)+4a\(^n\)+2a\(^n\)+8, next you will group them

(a\(^{2n}\)+4a\(^n\))+(2a\(^n\)+8). Notice you can factor out a few terms for them. Correct? Factor those terms out for me.

Answer 16

HINT: x\(^{2b}\)+x\(^b\) = x\(^b\)(x\(^b\)+1)....it's good to remember your exponent rules here.

Answer 17

NOTE: x\(^b\)(x\(^b\))= x\(^{b+b}\)= x\(^{2b}\)

Answer 18

\[a ^{2n}(6^{an}+8)\]a ^{2n+n}

Answer 19

No, not quite.

Answer 20

where did I make the mistake

Answer 21

\[6a ^{n}\]

Answer 22

Here's another HINT: a\(^n\)(a\(^n\)+4) that is ONE of them.

Answer 23

the second should be \[a ^{n}(a ^{n}+8)\]

Answer 24

No! You can ONLY factor out a 2 from the second one!

Answer 25

2\[2(a ^{n}+8)\]

Answer 26

actually IT SHOULD BE +4

Answer 27

YES! Correct!

Answer 28

So, now that you have everything factored, what is your final answer? I can compare it.

Answer 29

(__ + __)(__ +__) what are the __ ?

Answer 30

\[A ^{N}(A ^{n}+4) 2 ( a^{n}+4)\]

Answer 31

Not quite, you forgot the +2, but the final answer is: \(\sf\color{red}{(a^n+2)(a^n+4)}\)

Answer 32

thank you for your patience and help