Question

Solve for X : not looking for an Answer need to know how to approach the problem. I have an idea, but unsure if correct.

2x^3+6x^2-x=3

Answer 1

factor out x

Answer 2

2x^3+6x^2-x=3
Bring all the terms to one side
2x^3+6x^2-x-3=0
Nw try to factorise 2x^3+6x^2 and -x-3.
Can u tell me vat u get @immings?

Answer 3

2(x^3+3x^2) -(x+3)

Answer 4

2(x^3+3x^2) -(x+3)
U can factor out x^2 from x^3+3x^2.

Answer 5

ahh, forgot to do that. 2x^2(x+3) -(x+3)

Answer 6

so I can cancel out (x+3) and multiply (2x^2)(-1)?

Answer 7

No u can't cancel it. U hav to factorise it.
To make it simple let us assume
u=x+3.
2x^2(x+3) -(x+3)
2x^2u-u
Nw can u factorise it?

Answer 8

I don't see what I can factor out here.

Answer 9

an x maybe?

Answer 10

u can factor out u from 2x^2u-u.
Then u get
u(2x^2-1)
u=(x+3)
so (x+3)(2x^2-1). clear?

Answer 11

Yes. it makes more sense this way

Answer 12

Nw (x+3)(2x^2-1)=0
Either x+3=0 or (2x^2-1)=0
Can u find x nw?

Answer 13

Yes. Thank you. when I got to this step I was using LD to try to solve it. :s

Answer 14

oh k. yw:)