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)?
 A: 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
$$\begin{align}&3x^3-4x^2-x+2=3(x-1)^2\left(x+\frac 23\right)\\
\iff &3x^3-4x^2-x+2=(3x+2)(x-1)^2.\end{align}$$
A: This is a standard fact in Abstract Algebra, which goes by the name Root-Factor Theorem: given a polynomial $f(x)\in F[x]$ where $F$ is a field, $x-c$ is a factor of $f(x)$ if and only if $c$ is a root of $f(x)$.
A number $c$ is a root of $f(x)$ if and only if $f(c)=0$.
Since $f(1) = 3-4-1+2=0$, $1$ is a root of $f(x)$. So $x-1$ will divide $f(x)$ with zero remainder, i.e.,
do the long division for polynomials to obtain
$$f(x) = 3x^3-4x^2-x+2=(x-1)(3x^2-x-2).$$
By inspection or by using the quadratic formula, we see that $3x^2-x-2$ factors as $(3x+2)(x-1)$. So
$$
f(x) = 3x^3-4x^2-x+2= (x-1)(3x+2)(x-1).
$$
It is now clear what the $3$ roots of $f(x)$ are: $x=1, 1, -\displaystyle{\frac{2}{3}}$.
A: Here is a demo of the division of a polynomial by $x -\alpha$ using Horner's algorithm. I'll take the example $f(x)=3x^3-4x^2-x+2$ and $\:\alpha=7$.
We make a table with coefficients of the dividend in
decreasing degrees order in the first row:
$$\begin{array}{r}
\\ \\ \times 7\quad
\end{array}\begin{array}{|r|r|r|r|}
\hline
3\;\phantom{\nearrow} & -4\;\phantom{\nearrow} & -1\;\phantom{\nearrow} & 2 \\
\hline 
\downarrow \phantom{\nearrow}& 21\:\phantom{\nearrow} & 119\;\phantom{\nearrow} & 826 \\
\hline
3 \nearrow & 17\nearrow  & 118 \nearrow & \color{red}{828} \\
\hline
\end{array}$$
The result is that the quotient is the quadratic polynomial $\;q(x)=3x^2+17x+118$ and the remainder is $r=828$. Of course, if $\alpha$ is a root of the dividend, the remainder is $0$.
A: 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 $3x^3- 4x^2- x+ 2= (x- 1)(3x+ 2)(x- 1)$ 
$\rightarrow (x- 1)^2(3x+ 2)$
