Question

While I'm still logged in... can someone help me here: (z^3 + 4z^2 - 4z - 16) / (z + 2)

I won't lie, I'm way beyond this. I just really suck at algebra, and this is only a fraction of the problem, so I honestly don't feel all that guilty outsourcing.

Answer 1

z^2 (z+4) -4(z+4)

Answer 2

@joedwar Do you want to know how to factor the given expression, simplify, or what? I think it will factor.

Answer 3

I just need it in the simple (a^2 + a + 0) form so that I can finish the problem it is in.

Answer 4

From the work @shubhamsrg posted and picking up there.

z^2 (z+4) -4(z+4) =

Note that (z+4) is a common factor.

So, z^2 (z+4) -4(z+4) = (z+4)(z^2-4) =

(z+4) (z+2)(z-2) --> numerator

Answer 5

(z + 2) in the denominator takes out the (z+2) in the numerator.

Answer 6

That leaves (z-2)*(z+4) where z is not equal to -2.

Answer 7

@Directrix
So, z^3 + 4z^2 - 4z - 16 / (z + 2 = z + 4)^2 ?

Answer 8

Ah, I mean (z + 4)^2 sry

Answer 9

Do you get why z is not = -2? @joedwar

Answer 10

um

Answer 11

I don't need to know what z is or isn't.

Answer 12

I just need the equation's factors, with at least one of them being (z + 2)

Answer 13

You said that this problem is a part of a larger problem so the z cannot be -2 may be of importance. I don't know. It would be remiss not to mention it.

Answer 14

Ah, no. Want me to just past the whole equation for you so far?

Answer 15

If z = -2 then (z+2)/(z+2) = 1 becomes invalid because the division is 0/0 which is undefined.

Answer 16

If you like, post it.

Answer 17

((z^3 + 4z^2 - 4z - 16) / (z + 2)) / (z^2 + z - 12)

Answer 18

I just need to divide (z^3 + 4z^2 - 4z - 16) by (z + 2) and then simplify the equation.

Answer 19

Where is the MAIN division symbol?

Answer 20

((z^3 + 4z^2 - 4z - 16) / (z + 2)) = (z-2)*(z+4) from above.

Answer 21

\frac{ (z^3 + 4z^2 - 4z - 16) / (z + 2) }{ (z^2 + z - 12) }

Answer 22

well, that didnt work

Answer 23

\[\frac{ (z^3 + 4z^2 - 4z - 16) / (z + 2) }{ (z^2 + z - 12) }\]

Answer 24

There, that's what I have writen down.

Answer 25

All I need to do is bring this to simplest form.

Answer 26

[(z-2)*(z+4)]/ [(z^2 + z - 12)] is where we are.

So, factor (z^2 + z - 12) ---> Your task.

Answer 27

Um, I think that's wrong.

Answer 28

It wouldn't be (z-2)*(z+4). That doesn't even leave us with a z^3

Answer 29

Your job is to factor the following: factor (z^2 + z - 12).

The work looks okay to me so far.

Answer 30

so, bottom first, top later?

Answer 31

(z-3)(z+4)

Answer 32

(z^3 + 4z^2 - 4z - 16) / (z + 2)

We lost the z^3 when z divided into z^3 --> z^2 times.

Answer 33

I know that, but you wrote [(z-2)*(z+4)]/ [(z^2 + z - 12)], with only two factors on top, with no z^2

Answer 34

[(z-2)*(z+4)]/ [(z^2 + z - 12)] = [(z-2)*(z+4)]/ [(z-3)*(z+4)]

Answer 35

\[\frac{ (z^3 + 4z^2 - 4z - 16) / (z + 2) }{ (z-3)(z+4) }\]

Answer 36

That's where I'm at.
I'm pretty sure you lost a factor up top there.

Answer 37

The (z+4) factors divide out, provided z is not equal to -4.

Answer 38

>I'm pretty sure you lost a factor up top there.
Find where I lost it.

Answer 39

I'm so smart :D
I was thinking that the z+2 was the one from earlier xD

Answer 40

Even worse, I was reading it as z+2, not z - 2 ... I'm such an idiot

Answer 41

I think the best thing to do is to close this thread and begin a new one with the entire correct problem. We'll start over.

Answer 42

Okay, I'll mention you so you can find it.

Answer 43

Okay