How to factor this Polynomial

4x^3 - 18x^2 + 32 = 0 , and than find the roots, thanks.

...then you will be happy ?

4x^3 - 18x^2 + 32 = 0

common factor of 2...

2x^3 - 9x^2 + 16 = 0

[rational root theorem]

f(4) = 0, so (x - 4) is a factor

use synthetic division to find the quadratic factor...

4]..2.....- 9.....0.....16
_______8__- 4___-16
....2.....- 1...- 4.......0

(x - 4)(2x^2 - x - 4) = 0

check...read your textbook...practice...
THEN you'll be happy !

4x^3 - 18x^2 + 32 = 0
2 (2x^3 - 9x^2 + 16) = 0
2x^3 - 9x^2 + 16 = 0 => x = 4 is a root, divide by x - 4:
(2x^3 - 9x^2 + 16)/(x - 4) = 2x^2 - x - 4 hence:
=> (x - 4)(2x^2 - x - 4) = 0
x = 4
2x^2 - x - 4 = 0
x^2 - 1/2*x = 2
x^2 - 1/2*x + 1/16 = 2 + 1/16
(x - 1/4)^2 = 33/16
x - 1/4 = ±√33 / 4
x = 1/4*(1 ± √33)

... 4x^3 - 18x^2 + 32 = 0
or 2x^3 - 9x^2 + 16 = 0
or (x - 4) (2x^2 - x - 4) = 0
or x = { 4, (1/4)(1±√33) }

1) Take out a two to get 2(2x^3-9x^2+16)
2) Take out a (x-4) to get 2(x-4)(2x^2-x-4)

Final solution 2(x-4)(2x^2-x-4)