Factoring a polynomial I am trying to factor the following polynomial: 
$$ 4x^3 - 8x^2 -x + 2 $$
I am trying to do the following: $ 4x^2(x - 2)-x+2 $ but I am stuck.
Thanks for your help.
edit: correction.
 A: You can factor as
$$
4x^2(x-2)-(x-2)=4x^2(x-2)-(1)(x-2)
$$
Then factor out the $x-2$ to get
$$
(x-2)(4x^2-1).
$$
But, you may further factor $4x^2-1$ to get
$$
(x-2)(2x-1)(2x+1).
$$
A: I'll use your polynomial to illustrate a more general procedure for factoring polynomials with integer coefficients (and assuming it has at least one rational root):
First, guess a root  of $4x^3-8x^2-x+2$. The so called  "rational roots test"  will be helpful here.
Eventually, you'll discover that $x=2$ is a root of $4x^3-8x^2-x+2$.  This will imply that your polynomial has the form
$$
\tag{1}(x-2)(ax^2+bx+c),
$$
for some constants $a, b, c$.
To find those constants, you could do one of two things (and maybe more)


*

*perform the division $4x^3-8x^2-x+2\over x-2$.

*expand (1) and set it equal to the original polynomial.  Setting the coefficients of the two sides of this equation equal to each other will give you a system of equations that are solvable for $a$, $b$, and $c$.


Once you've figured out what $a,b$, and $c$ are, factor the quadratic.

Using method (2), we have:
$$
4x^3-8x^2-x+2 = ax^3+(b-2a)x^2+(c-2b)x-2c
$$
A moment's reflection reveals that $c=-1$; whence $b=0$; whence $a=4$. Thus
$$\eqalign{
4x^3-8x^2-x+2 &= (x-2)(4x^2-1)\cr
 &= (x-2)(2x+1)(2x-1).\cr
}
$$
Of course, the other answers are more suitable to your problem; but in the event that your polynomial doesn't factor nicely (such as for $x^3+6x^2+11x+6$), you might try using this approach.
A: Franco, you said you are stuck.  Did you have an approach?
In general, for a cubic equation to be factored means the person who is asking this question is indirectly giving you a hint that there is at least one real root.  
Your approach should be as follows:
Step 1:
first to try to see if x=0, 1 -1, 2 or -2 is a root.  Chances are in some cases that one of these is a root.  In that case, let us say you figured that $x=2$ is a root of $4x^3 - 8x^2 -x + 2$, then if you know factoring polynomials a little bit then you could immediately figure $4x^3 - 8x^2 -x + 2 = (x-2)(4x^2-1)$ which means the other factors can be found by the fact that $(4x^2-1^2) = (2x-1)(2x+1)$.
If in some cases you cannot find the obvious root as described in Step 1, then 
Step 2:
Observe if there is a pattern like $4x^3-8x^2$ has two coefficients $4$ and $8$, which has the same pattern as $-x+2$, i.e. $4x^3-8x^2 = 4x^2(x-2)$ and $-x+2 = -(x-2)$, which means you can combine those as 
$$(x-2)4x^2-(x-2) = (x-2)(4x^2-1)$$  
And in some cases as here you could also do $(4x^3-x) - (8x^2-2)$ which gives you
$$ x(4x^2-1)-2(4x^2-1) = (x-2)(4x^2-1) = (x-2)(2x-1)(2x+1)$$
Luckily with this question, Step 1 and Step 2 worked.  What if you have a cubic equation that does not have obvious ways to factor.
Step 3:   For a general cubic equation $ax^3+bx^2+cx+d=0$, apply the substitution
$$ x = y - \frac{b}{3a} $$
then we get 
$$ a\left(y-\frac{b}{3a} \right)^3 + b\left(y-\frac{b}{3a} \right)^2+c\left(y-\frac{b}{3a} \right)+d = 0 $$
which simplifies to
$$ ay^3 + \left( c-\frac{b^2}{3a}\right)y+ \left(d + \frac{2b^3}{27a^2} - \frac{bc}{3a} \right) = 0 $$
This is called a depressed cubic equation, because the square term is eliminated.  It is much easier to use this and then
find the roots. (back substitute to get the roots in terms of $x$)
For example $$2x^3-18x^2+46x-30=0$$
Substitute $ x=y+3$ and simplify this cubic equation to
$2y^3-8y=0 \Rightarrow y=0,2,-2$ which then gives the roots as
$x=1,3, $ and $5$.
A: $(4x^3-x)-(8x^2-2)=x(4x^2-1)-2(4x^2-1)=(x-2)(4x^2-1)=$
$=(x-2)(2x-1)(2x+1)$
