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)?

Yes, that is correct. If $\alpha$ is the root of the polynomial $P(x)$, then $x-\alpha \mid P(x)$.
Also, we have several methods to factorise your polynomial:

*

*Polynomial long division


*Synthetic division
Finally, I can suggest the following general method:
$$\begin{align}&3x^3-4x^2-x+2=(x-1)(3x^2+ax+b)\\
\implies &3x^3-4x^2-x+2=3x^3+x^2(a-3)-x(a-b)-b\\
\implies &\begin{cases}a-3=-4\\ a-b=1\\ b=-2\end{cases}\\
\implies &(a,b)=(-1,-2)
\end{align}$$
Then applying the quadratic formula we get,
$$3x^2-x-2=(3x+2)(x-1)$$
Therefore, we have
$$3x^3-4x^2-x+2=(3x+2)(x-1)^2$$

Small supplement:
Since $f'(x)=9x^2-8x-1$ and $f'(1)=0$, then we see that $x_1=1$ and $x_2=1$ are repeated root of the polynomial $f(x)=3x^3-4x^2-x+2$.
This implies,
$$(x-1)^2\mid 3x^3-4x^2-x+2$$
Then, Vieta's formulas tells us,
$$\begin{align}1\times 1\times x_3=-\frac 23\implies x_3=-\frac 23\end{align}$$
Therefore, we conclude that
$$\begin{align}&3x^3-4x^2-x+2=3(x-1)^2\left(x+\frac 23\right)\\
\iff &3x^3-4x^2-x+2=(3x+2)(x-1)^2.\end{align}$$
