Factor Teorem: Solving the equation 3x^3+7x^2+22x-8=0

[MissT]

This question has three parts. i completed the first two parts but i'm stuck at part three. The question is as follow:

Given that (x-2) is a factor of 3x^3+7x^2+hx-8:
(i) Show that the value of h= -22
Solution: Since (x-2) is a factor, then f(2)=0 therefore substitute 2 wherever x is in the equation
3(2)^3+7(2)^2=h(2)-8
=24+28-8+2h
=2h+44
=2h= -44
=-44/2
=-22

(ii) Factorise completely 3x^3+7x^2+22x-8
I used synthetic division and got 3x^2+13x-4 when factored i got (x-2)(3x-1)(x+4)

(iii) Hence or otherwise solve: 3x^3+7x^2+22x-8=0 This is where i'm stuck. I was never good at equations so i need some help with solving part (iii) of this question. Thanks

[stapel_eliz]

(ii) Factorise completely 3x^3+7x^2+22x-8

...i got (x-2)(3x-1)(x+4)

(iii) Hence or otherwise solve: 3x^3+7x^2+22x-8=0

Take the factored form, set it equal to zero, and then solve each of the three linear factors:

. . . . .x - 2 = 0

. . . . .3x - 1 = 0

. . . . .x + 4 = 0

[MissT]

Thank you stapel_eliz