Question

More factoring.

Answer 1

I haven't a clue on this one.

Answer 2

@Hero @Astrid1

Answer 3

@snowflake0531

Answer 4

xddd, maybe it's unfactorable? hehe

Answer 5

you want me to do all that?
xdxdddd
do you know how to do any of it?

Answer 6

oh waitttttttt, nevermind

Answer 7

YOu shoulda gave me the top part xd

Answer 8

i was gonna figure out the first one and try it on my own after that

Answer 9

oh, i found the notes...

Answer 10

2

Answer 11

but yes, it's 2 xd

but i wrote it wrong earlier

Answer 12

but how doo you cube root x^3

Answer 13

So now that we know

\[x^3 + 2^3\]

Answer 14

im still confused as to how that happened...

Answer 15

ohhhhhh, 2^3 is equal to 8, so you can plug in 2^3 as 8

Answer 16

ok

Answer 17

ok, thx

Answer 18

\[x^2 + y^2 = (x+y)(x^2 -xy + y^2 )\]

Answer 19

So you got it?

Answer 20

wouldnt it be x^3 + y^3?

Answer 21

x is still x
but plug in y as 2

Answer 22

yea, so x^3 + 2^3

Answer 23

\[x^3 + 2^3 = (x+2)(x^2 -2x +2^2 )\]

Answer 24

it confused me cuz u put x^2 + y^2

Answer 25

i see now

Answer 26

ye, for some reason i keep on typing wrong lmao

Answer 27

lollz

Answer 28

Btw, you don't use this for every single one
for example, looking at the last one, you do factor by grouping
do you know how to do that?

Answer 29

its in the notes, ill see if I can get it on my own.

Answer 30

so the answer to this one would be (x + 2)(x^2 - 2x + 4), right?

Answer 31

yea