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 : $1-3x+4x^3=(1-2x)^2(1+x)$ ?

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A number $\alpha$ is a root of multiplicity $\geq 2$ of a polynomial $P(x)$, if and only if $P(\alpha)=P'(\alpha)=0$. – 1015 Feb 9 '13 at 12:30
Perform the long division $(1-3x+4x^3)/(x+1)$. – David Mitra Feb 9 '13 at 12:37
Yes, both verbs can be used. 'Factorize' and 'factor' (when used as a verb) mean the same thing. – Alraxite Feb 9 '13 at 13:45
Thank you very much. I will remember that "factor" and "fatorize" mean the same thing. – user61594 Feb 9 '13 at 14:14

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.

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I chose the simplest answer but the other ones are interesting too. – user61594 Feb 9 '13 at 17:04

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 $(1-2x)^2$

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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.

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What's the definition of the gcd of $f$ and $f^{'}$ ? Why do you express $f$ with the help of $f^{'}$ ? I don't understand :( – user61594 Feb 9 '13 at 12:46
Could you please give more explanations ? Thank you very much. – user61594 Feb 9 '13 at 14:16

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 $$\pm 1, \quad \pm \frac{1}{2}, \quad \pm \frac{1}{4}.$$

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There's a formula for equations of degree 3, but in the practice Ruffini's rule is used.

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You should include more details, otherwise this is a comment, not an answer. – Julian Kuelshammer Feb 9 '13 at 13:00