Question

Factor and find all solutions of 3x^3 - 6x^2 - 9x = 0

Answer 1

\[3x^3 - 6x^2 - 9x = 0\]

Answer 2

you can start off by factoring out the GCF 3x


3x^3 - 6x^2 - 9x = 0

3x(x^2 - 2x - 3) = 0


now factor x^2 - 2x - 3

Answer 3

So that would be
3x(x-1)^2 = 3

Answer 4

no you can't move that 3 over like that

Answer 5

and x^2 -2x doesn't factor to (x-1)^2

Answer 6

I used the completing the square method...

Answer 7

factor x^2 - 2x - 3

find two numbers that multiply to -3 and add to -2

Answer 8

I don't think there r any...

Answer 9

This is how i did it,

\[3x^3-6x-9x=0\]
\[3x(x^2-2x-2)=0\]
\[3x(x^2-2x+1)= 2+1\]
\[3x(x^2-2x+1)=3\]
\[3x(x-1)^2 =3\]
\[3x(x-1)=\pm \sqrt{3}\]
\[3x(x)=1\pm \sqrt{3}\]

Answer 10

Then i got stuck there

Answer 11

sure there are: -3 and +1

Answer 12

-3 + 1 = -2

-3 * 1 = -3

Answer 13

since the two numbers are -3 and 1, this means x^2 - 2x - 3 factors to (x-3)(x+1)

Answer 14

3x^3 - 6x^2 - 9x = 0

3x(x^2 - 2x - 3) = 0

3x(x-3)(x+1) = 0

now use the zero product property

Answer 15

but how did u get the 3 because u get a radical if you find the square root of 3

Answer 16

in your steps, you factored 3x^3 - 6x^2 - 9x incorrectly

Answer 17

you should get 3x(x^2 - 2x - 3)

NOT 3x(x^2 - 2x - 2)

Answer 18

also, I'm not completing the square. I'm simply factoring x^2 - 2x - 3

Answer 19

ooooohhhhhhh

Answer 20

Thanks lol it was just a silly mistake

Answer 21

you're welcome