Factoring a third degree polynomial with a given root My professor gave us the following polynomial:
$f(x) = 3x^3-4x^2-x+2$
Given is that $x = 1$ is a root of this function. We are asked to find the other ones.
He then told us that, given $x=1$ is a root, we now know that we can factorize this polynomial into $(x-1)$, and a second factor starting with $3x^2...$.
I'm familiar with the process of factoring polynomials, and of finding the roots by setting each factor to zero. But what are you actually doing when you factor a polynomial? In my mind it's just a formulaic way to find zeros of such functions. That's why I was a little confused when we are asked to do the reverse; finding the factor of a given root. More generally, if you know a root of a polynomial, for example $x=7$, can you then always conclude that one factor is $(x-7)$?
How should I proceed finding the other factor(s)?

Factoring a polynomial is basically the same as factoring an integer into primes. $34= 2(17), 36= 2^2(3^2$). Any polynomial can be factored, over the real numbers, into either linear terms, such as "$(x- a)$" where $a$ is a real number, or a irreducible quadratic term. (quadratics that cannot be factored further with real numbers)
Any polynomial can be factored, over the Complex numbers, into all linear factors.
The fact that $x = 1$ is a root of this equation means that $x - 1$ is a factor of $f(x)= 3x^3- 4x^2- x+ 2$.
So we can write $f(x) = (x- 1)(ax^2+ bx+ c) = ax^3+ (b- a)x^2+ (c- b)x- c$.   So we must have $a= 3, b- a= b- 3= -4$.
So $b= -1$, and $c- b= c+ 1= -1$ so $c= -2$.
That is $3x^3- 4x^2- x+ 2= (x- 1)(3x^2- x- 2)$.
Now, can $3x^3- x- 2$ be factors (with integer coefficients)?  If it did one factor would have to have $3x$ and the other $x$ (so that they multiply to $3x^2$) and one factor would have to have $-1$ and the other $2$ or one would have to have $1$ and the other $-2$.
Putting those together gives four possibilities:- 
To decide which, just multiply them.
$(3x- 1)(x+ 2)= 3x^2+ 5x- 2$ 
$(3x+ 2)(x- 1)= 3x^2- x- 2$ 
$(3x+ 1)(x- 2)= 3x^2- 5x- 2$ 
$(3x- 2)(X+ 1)= 3x^2+ x- 2.$ 
But, $3x^2- x- 2= (3x+ 2)(x- 1).$
So $3x^3- 4x^2- x+ 2= (x- 1)(3x+ 2)(x- 1)$ 
$\rightarrow (x- 1)^2(3x+ 2)$
