8
$\begingroup$

In one of the mathematics book, the author factorized following term

$$x^3 - 6x + 4 = 0$$ to

$$( x - 2) ( x^2 + 2x -2 ) = 0.$$

How did he do it?

$\endgroup$
  • 2
    $\begingroup$ If there is a cube term, it is not a quadratic. Also, a method for finding divisors of your polynomial, look at the factors of the constant term. $\endgroup$ – Edward Evans Jan 28 '17 at 19:52
  • 2
    $\begingroup$ $x=2$ is a root of this equation. Then by polynomial division of $\frac{x^3-6x+4}{x-2}$ we obtain $x^2+2x-2$. $\endgroup$ – projectilemotion Jan 28 '17 at 19:53
  • $\begingroup$ Do you know the Rational Root Test? $\endgroup$ – Bill Dubuque Jan 28 '17 at 19:54
  • 3
    $\begingroup$ A common first step for introductory problems is to guess and check certain integers close to zero to see if it will equal zero. Zero is the easiest to check because that would mean the constant term is zero, thats not the case here. $1$ doesn't work because that would be $1-6+4=-1\neq 0$. $2$ happens to work since this would be $2^3-6\cdot 2 + 4 = 8-12+4=0$. Since $2$ works, we know that the equation can be factored as $(x-2)q(x)$ where $q(x)=\frac{x^3-6x+4}{x-2}$. In general this won't always work, especially if the roots aren't even integers. Cardano's formula would help then. $\endgroup$ – JMoravitz Jan 28 '17 at 19:54
  • $\begingroup$ @BillDubuque No $\endgroup$ – Govinda Sakhare Jan 28 '17 at 19:56
6
$\begingroup$

There is a neat trick called the rational roots theorem. All we have to do is factor the first and last numbers, put them over a fraction, and take $\pm$. This gives us the following possible rational roots:

$$x\stackrel?=\pm1,\pm2,\pm4$$

due to the factorization of $4$. Checking these, it is clear $x=2$ is the only rational root, since

$$\begin{align}0&\ne(+1)^3-6(+1)+4\\0&\ne(-1)^3-6(-1)+4\\\color{#4488dd}0&=\color{#4488dd}{(+2)^3-6(+2)+4}\\0&\ne(-2)^3-6(-2)+4\\0&\ne(+4)^3-6(+4)+4\\0&\ne(-4)^3-6(-4)+4\end{align}$$

leaving us with

$$x^3-6x+4=(x-2)(\dots)$$

We can find the remainder through synthetic division:

$$\begin{array}{c|c c}2&1&0&-6&4\\&\downarrow&2&4&-4\\&\hline1&2&-2&0\end{array}$$

which gives us our factorization:

$$x^3-6x+4=(x-2)(x^2+2x-2)$$

$\endgroup$
  • $\begingroup$ OP does not know RRT - see the comments. $\endgroup$ – Bill Dubuque Jan 28 '17 at 19:59
  • $\begingroup$ @BillDubuque Oh. Then give me a moment $\endgroup$ – Simply Beautiful Art Jan 28 '17 at 20:00
  • $\begingroup$ @SimplyBeautifulArt Why 1,2 and 4. Why are we not checking equation with value 3? it has anything to do with c term ax^3+bx+c ? $\endgroup$ – Govinda Sakhare Jan 28 '17 at 20:09
  • $\begingroup$ @piechuckerr By the rational roots theorem, I need only check the factors of the last constant if the leading coefficient is $1$. $3$ does not divide into $4$, so I needn't check it. $\endgroup$ – Simply Beautiful Art Jan 28 '17 at 20:10
  • $\begingroup$ @SimplyBeautifulArt i.e if constant term is 6 then 1,2,3 and 6 right? $\endgroup$ – Govinda Sakhare Jan 28 '17 at 20:12
4
$\begingroup$

Since you do not know the Rational Root Test, let's consider a simpler case: the Integer Root Test.

If $\,f(x)= x^3+6x+4\,$ has an integer root $\,x=n\,$ then $\,n^3+6n+4 = 0\,$ so $\,(n^2+6)\,\color{#c00}{n = -4},\,$ hence $\,\color{#c00}{n\ \ {\rm divides}\ \ 4}.\,$ Testing all the divisors of $4$ shows that $2$ is root, $ $ hence $\,x-2\,$ is a factor of $f$ by the Factor Theorem. The cofactor $\,f/(x-2)\,$ is computable by the Polynomial (long) Division algorithm (or even by undetermined coefficients).

Remark $\ $ This is a very special case of general relations between the factorization of polynomials and the factorizations of their values. For example, one can derive relations between primality and compositeness of polynomials based on the same properties of their values. For example, since $\ 9^4\!+8\ $ is prime so too is $\, x^4+8\,$ by Cohn's irreducibility test. See this answer and its links for some of these beautiful ideas of Bernoulli, Kronecker, and Schubert.

$\endgroup$
  • $\begingroup$ brilliant, liked every bit of it, Sadly I can not unmark other answer and select this one coz that too is elegant answer. $\endgroup$ – Govinda Sakhare Jan 29 '17 at 16:50
  • 4
    $\begingroup$ @piechuckerr No problem (i don't care about acceptance, votes, etc, only about sharing beautiful mathematics). $\endgroup$ – Bill Dubuque Jan 29 '17 at 17:05
1
$\begingroup$

Note: I understand that there is already an accepted answer for this question, so this answer may be useless, but regardless, I'm still posting this to spread knowledge!

A simple way to factorize depressed cubic polynomials of the form$$x^3+Ax+B=0\tag1$$

Is to first move all the constants to the RHS, so $(1)$ becomes$$x^3+Ax=-B\tag2$$ Now, find two factors of $B$ such that one fact minus the square of the other factor is $A$. We'll call them $a,b$ so$$\begin{align*} & a-b^2=A\tag3\\ & ab=-B\tag4\end{align*}$$ Multiply $(2)$ by $x$, add $b^2x^2$ to both sides and complete the square. Solving should give you a value of $x$ and allow you to factor $(1)$ by Synthetic Division.


Examples:


  1. Solve $x^3-6x+4=0$ (your question)

Moving $4$ to the RHS and observing its factors, we have $-2,2$ as $a,b$ since$$-2-2^2=A\\-2\cdot2=-4$$Therefore, we have the following:$$x^4-6x^2=-2\cdot2x$$$$x^4-6x^2+4x^2=4x^2-4x$$$$x^4-2x^2=4x^2-4x$$$$x^4-2x^2+1=4x^2-4x+1\implies(x^2-1)^2=(2x-1)^2$$$$x^2=2x\implies x=2$$ Note that we do have to consider the negative case when square rooting, but they lead to the same pair of answers. So it's pointless.

  1. Solving $x^3+16x=455$

A factor of $455$ works, namely when $a=65,b=7$.$$65-7^2=16$$$$65\cdot7=455$$ Therefore,$$x^4+16x^2=65\cdot7x$$$$x^4+65x^2=49x^2+455x$$$$\left(x^2+\dfrac {65}{2}\right)^2=\left(7x+\dfrac {65}{2}\right)^2$$$$x=7$$

$\endgroup$
0
$\begingroup$

The rational root theorem gives a list of all possible rational roots of a polynomial with integer coefficients that have a given leading coefficient and a given constant coefficient. In this case, the leading coefficient is $1$ and the constant coefficient is $4.$ The theorem tells us that all rational roots are in the set $\left\{ \pm\dfrac 1 1, \pm\dfrac 2 1, \pm \dfrac 4 1 \right\},$ the numerator being in this the only divisor of the leading coefficient $1$ and the denominators being the divisors of the constant coefficient $4$. That doesn't mean there are rational roots; it only means there are not any that don't belong to this set. There are only six members of this set, so it's easy to plug in all of them and see if you get $0$. When you plug in $2$, you get $0$, so there's your factorization.

$\endgroup$
  • $\begingroup$ OP does not know RRT - see the comments. $\endgroup$ – Bill Dubuque Jan 28 '17 at 19:59
  • 1
    $\begingroup$ @BillDubuque : Fortunately I linked to the Wikipedia article about it. $\endgroup$ – Michael Hardy Jan 28 '17 at 19:59
  • 1
    $\begingroup$ @BillDubuque : That he doesn't recognize that name of the theorem doesn't mean he doesn't know the theorem. He could know it by a different name. $\endgroup$ – Michael Hardy Jan 28 '17 at 20:01
  • $\begingroup$ I thought the comment(s) might nudge someone to give an exposition on this simpler integer case. But no one did, so I added an answer doing so. $\endgroup$ – Bill Dubuque Jan 28 '17 at 20:40
-1
$\begingroup$

If $P$ is a polynomial with real coefficients and if $a\in\mathbb{R}$ is a root, which means that $P(a)=0$, then there exists a real polynomial $Q$ such that $\forall x\in\mathbb{R},\quad P(x)=(x-a)\,Q(x)$.

On this case, you can see by inspection, that $P(2)=0$.

It remains to find real constants $A,B,C$ such that :

$$\forall x\in\mathbb{R},\quad x^3-6x+4=(x-2)(Ax^2+Bx+C)$$

Identification of coefficients leads to $A=1$, $-2C=4$ and, for example, $A-2B=0$ (equating the coeffts of $x^2$ in both sides).

$\endgroup$
  • $\begingroup$ Since the OP doesn't know RRT, the answer I posted earlier seems to me a (not so bad) - one ! It's clear that one can answer at a higher level, but it's seems to me important to answer at a level compatible the mathematical background of the OP. I have the feeling that, if he (or she) had known the RRT, he would certainly never have asked that question. $\endgroup$ – Adren Jan 28 '17 at 20:11
  • $\begingroup$ Actually we don't need the full RRT here, only a simpler integer case, e.g. see my answer. $\endgroup$ – Bill Dubuque Jan 28 '17 at 20:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.