Polynomial factoring $1-3x+4x^3$ I want to factorize (or factor ? can both verbs be used ?) $1-3x+4x^3$. I notice that $\frac{1}{2}$ and $-1$ are roots of the polynomial.
My questions are : 
1) how do you notice that $\frac{1}{2}$ is "two times" a root of this polynomial ?
2) what is the quickest way to arrive to this conclusion : $1-3x+4x^3=(1-2x)^2(1+x)$ ?
 A: Just Differentiate the polynomial w.r.t $x$ and then check whether $0.5$ is a root of this as well.If it is  then it is $0.5$ is "two times" a root of this polynomial.
As you know that $-1$ is a root of this polynomial so just factotise the polynomial with $(x+1)$ as one of its factor. then the other factor will be $(1-2x)^2$
A: You can detect multiple roots of $f$ by computing the gcd of $f$ and $f'$.
If you have already found the root $\frac12$, you can of course simply try to divide $f$ by $(x-\frac12)$ repeatedly ...:
$$f(x) = 4x^3-3x+1,\quad f'(x)=12x^2-3=3(4x^2-1)$$
$$\implies \quad f(x)=x\cdot \frac13f'(x)-2x+1$$
hence any multiple root of $f$ is a root of $-2x+1=0$, i.e. $x=\frac12$. As it is indeed a root, divide $f(x)$ by $(2x-1)^2$ to obtain $(x+1)$ as remaining factor.
A: By observation $f(x)=4x^3-3x+1$ has $-1$ as a zero.
From this, we see that $f(x)=(x+1)(4x^2+kx+1)$ for some $k$. Solving for $k$ by equating coefficients gives $k=-4$, so that $f(x)=(x+1)(4x^2-4x+1)=(x+1)(2x-1)^2$.
This then gives us the fact that $\frac12$ is a double zero.
A: There is a useful elementary result, that tells you the following.
Suppose
$$
a = a_{n} x^n + a_{n-1} x^{n-1} + \dots + a_1 x + a_{0}
$$
is a polynomial with integer coefficients, and $a_n \ne 0$. 
Suppose $a$ has a rational root $u/v$, with $u$ and $v$ coprime integers.
Then $u \mid a_{0}$ and $v \mid a_{n}$.
In your case, this tells you that the possible rational roots are
$$
\pm 1, \quad \pm \frac{1}{2}, \quad \pm \frac{1}{4}.
$$
A: There's a formula for equations of degree 3, but in the practice Ruffini's rule is used.
