Question

Need to factor t^2+4z^2

Answer 1

do you know complexx numbers ?!

Answer 2

No I don't

Answer 3

If this what you mean
\[t^2+4z^2\]
it can't be factorized

Answer 4

Please let me know if this is correct:
(t+2z)(t+2z) factor completed

Answer 5

in other words it would be written prime... correct?

Answer 6

Sorry , but I didn't understaand what you really want to do ! :)

Answer 7

ok I have another one for you. I am asked to factor this polynomial: y^3+y^2-4y-4

Answer 8

okk ! this ! it can be factorized :) ! you may know that\[(-1)^3+(-1)^2-4(-1)-4=0\]
so , this "polynomial" can be factorized by (y--1)=(y+1)

Answer 9

what would be considered the common factor group?

Answer 10

kk you didn't undersand what I said ! It's normal ! we will try to do this in other way ! kk ?!

Answer 11

This is what I has before:

y^3+y^2-4y-4
y(y^2+y) -4(y+1)
(y^2+y)-(y-4)
(y^2+y)(y-4)(4+-1)

Answer 12

ok. trust me it has been a reeeallllyyy long time since I have done algebra and my brain is hurting over this lol

Answer 13

I have to be able to show my steps as well as explain it. At this point I am just trying to figure out how to do this :)

Answer 14

no no ! good ! you were in the right way ! \[y^3+y^2=y^2(y+1)\]
-4y-4=-4(y+1)

Answer 15

you're reallyy a smart person !

Answer 16

I seriously doubt that right now lol

Answer 17

no no ! you're doing good trust me !

Answer 18

How would I go about explaining how I arrived at this conclusion?

Answer 19

so would my common factor be the (y+1)?

Answer 20

yeaaaaah !

Answer 21

now give me the final answer !

Answer 22

(y+1)(y+1)(y+-4)

Answer 23

I have to utilize grouping because of the four terms right

Answer 24

give me another answer :'( !

Answer 25

\[y^3+y^2-4y+4=y^2(y+1)-4(y+1)\]
now (y+1) is the common factor ! kk

Answer 26

go on what are waiting for ?! :) :)

Answer 27

differences in squares would be: (y+1)(y^2+4)?

Answer 28

(y+1)(y+2)(y+-2)?

Answer 29

completely factored at (y+1)(y^2+2)(y+-2)

Answer 30

now you're good ! But you wrote (y^2+4) correct this
the final ansswer is correct !

Answer 31

so is (y+1)(y^2+2)(y^-2) correct?

Answer 32

yeaaaah that's correct !

Answer 33

thank you !!!! :)

Answer 34

You're welcome !anytime !

Answer 35

o wait shouldn't it be like this:
(y+1)(y+2)(y-2)

Answer 36

final factorization that is?

Answer 37

(y+1)(y+2)(y+-2) (Ugh)

Answer 38

okk I didn't see well !
it should be !
the right answer is (y+1)(y+2)(y-2) which is the same as (y+1)(y+2)(y+-2)

Answer 39

oh ok lol

Answer 40

thank you :D

Answer 41

yw ! :D !

Answer 42

and good luck !

Answer 43

thank you because boy am I going to need it

Answer 44

:) yeaaah we all need luck !