Question

Solving equation help please???

Answer 1

I tested every answer here and none of them were equal to zero, what am I doing wrong?

Answer 2

We could factor by grouping..

Answer 3

\(\ 9x^3 +9x^2-4x-4\)
Do you know how to factor by grouping? @JoeJoldin

Answer 4

Sort of, will the next step look like this?:

x^3 + x^2 - 4 - 1

Answer 5

Umm. no

Answer 6

\(\ 9x^3 +9x^2-4x-4\)
We would seperate the first 2 terms and facotr them and sperate the last two terms and factor them..

Answer 7

\(\ 9x^3 +9x^2~~~~~~~~~~~~~-4x-4\)

\(\ 9x^2(x+1) ~~~~-4x-4\)
and then we factor the other half.

Answer 8

not sure how to do that

Answer 9

-4x-4
What is common?

Answer 10

4

Answer 11

More like -4

Answer 12

And then we will get

\(\ 9x^2(x+1) ~-4(x+1)\)

Answer 13

And if we take out (x+1) from it we can get it factored!

Answer 14

So now its just 9x^2-4?

Answer 15

Umm more like

\(\ (x+1)(9x^2-4)\)

Answer 16

And we just set each one equal to 0

Answer 17

So what do I do now

Answer 18

\(\ x+1=0\)
and
\(\ x^2-4=0\)

Answer 19

My answer must be B then right?

Answer 20

Well there is no correct answer...

Answer 21

All of those answer choices are wrong..

Answer 22

yeah this has got me lost

Answer 23

And I am pretty sure I did everything correct.

Answer 24

@ParthKohli @Jhannybean Can check for me :P

Answer 25

Plugging in choices now.. And A isn't correct...

Answer 26

\[9x^2 - 4 = 0\]\[x+1 = 0\]

Answer 27

Yeah got that too.

Answer 28

I am getting B as the answer

Answer 29

\[(3x + 2)(3x - 2) = 0\]\[x+ 1 = 0\]

Answer 30

Oh wait B is the answer.

Answer 31

I should go to sleep now..