Factor theorem/ polynomial divison help needed!

Factor theorem/ polynomial divison help needed!?
could anyone please show me the full steps showing how to factorise completely : 2x^3+3x^2-18x+8
thanks.
(first time doing a question like this, so not sure what to do).

By the Rational Root Theorem, any rational root to 2x^3 + 3x^2 - 18x + 8 must be in the form ±p/q where p is a factor of 8 (the constant term) and q is a factor of 2 (the leading coefficient).

The factors of 8 are 1, 2, 4, and 8 and the factors of 2 are 1 and 2, so we have:
±p/q = ±1, ±2, ±4, ±8, ±1/2.

Going through this list, x = -4, x = 1/2, and x = 2 are all roots to this equation (and they are the only ones by the Fundamental Theorem of Algebra). Thus, by the Factor Theorem, x + 4, x - 1/2, and x - 2 must be factors. The product of these factors is (x + 4)(x - 1/2)(x - 2), but this product has a x^3 coefficient of 1, not 2. To fix this, introduce a factor of 2 to get:
2(x + 4)(x - 1/2)(x - 2) = (x + 4)(2x - 1)(x - 2).

Therefore, 2x^3 + 3x^2 - 18x + 8 = (x + 4)(2x - 1)(x - 2).

I hope this helps!

we need to find root. let's try 2

2(2)^3 + 3(2)^2 - 18(2) + 8 =

2(8) + 3(4) - 36 + 8 =

16 + 12 - 28 =

28 - 28 = 0 so 2 is a root and can be written as (x - 2) now do division

.....................2x^2 + 7x - 4
.....................2x^3 + 3x^2 - 18x + 8
x - 2..............2x^3 - 4x^2
................................7x^2 - 18x
................................7x^2 - 14x
......................................… 4x + 8
......................................… 4x + 8

so we have as factors (x - 2)(2x^2 + 7x - 4). we then can factor the second term as (2x - 1)(x + 4) so we get

(x - 2)(2x - 1)(x + 4)