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I want to know other ways of factorization to get quadratic factors in this polynomial:

$$x^4+2x^3+3x^2+2x-3$$

Thanks in advance for your suggestions.

The original polynomial is $$x^6-2x^3-4x^2+8x-3$$ where the two found factors are $(x+1)$ and $(x-1)$ by synthetic division.

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    $\begingroup$ What do you mean by "other ways"? Which ones don't you want us to tell you about? $\endgroup$ – YukiJ Jan 3 '17 at 16:55
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    $\begingroup$ The factorization is $$(x^2+x-1)(x^2+x+3)$$ but I didn't find it by hand $\endgroup$ – Peter Jan 3 '17 at 16:58
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    $\begingroup$ @YukiJ "Other ways" out of: identifying sums and differences of cubes and perfect squares or synthetic division. $\endgroup$ – Isai Jan 3 '17 at 17:15
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    $\begingroup$ If you know the roots, then you can also factorize it. $\endgroup$ – YukiJ Jan 3 '17 at 17:16
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    $\begingroup$ A quartic is never irreducible over real polynomials. It always has at least one factorization into two quadratic factors. Finding those factors is the tricky part. $\endgroup$ – David K Jan 3 '17 at 19:22
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Note that $$(x^2 + x + 1)^2 = x^4 + 2x^3 + 3x^2 + 2x + 1$$which means that your polynomial is equal to $$ (x^2 + x + 1)^2 - 4 = (x^2 + x + 1)^2 - 2^2 = (x^2 + x + 1 - 2)(x^2 + x + 1 + 2) $$

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    $\begingroup$ If the downvoting party could please tell me what I did to deserve it, maybe I could actually make this answer (and future ones) better. As it stands now, I have learned nothing from the downvote, and intend to carry on exactly as before. $\endgroup$ – Arthur Jan 3 '17 at 17:35
  • $\begingroup$ +1 from me. What led you to try that first quadratic squared? Did you notice some pattern? $\endgroup$ – JoeTaxpayer Jan 3 '17 at 19:32
  • $\begingroup$ @JoeTaxpayer As with many problems like these, it relied partly on my having seen something like it before. More generally, we have $$(x^n + x^{n-1} + \cdots + x + 1)^2 = x^{2n} + 2x^{2n-1} + \cdots + nx^{n+1} + (n+1)x^n + nx^{n-1} + \cdots +2x + 1 $$ $\endgroup$ – Arthur Jan 3 '17 at 20:13
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    $\begingroup$ Related to the wonderful Demlo number $111^2=12321$ (take $x$ as the base of your positional numeral system). $\endgroup$ – Jeppe Stig Nielsen Jan 3 '17 at 23:42
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    $\begingroup$ @Arthur: downvote for no reason is getting very common here. I really like this answer. For me it is a mix of experience and felling. $\endgroup$ – Arnaldo Jan 4 '17 at 1:37
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One may try to use the rational root theorem. But that is not possible in this case because there is no rational root.

Another thing we can try (if we don't "see" anything) is write

$$(x^2+ax+b)(x^2+cx+d)\equiv x^4+2x^3+3x^2+2x-3$$

$$x^4+(a+c)x^3+(d+ac+b)x^2+(ad+bc)x+bd\equiv x^4+2x^3+3x^2+2x-3$$

and then solve:

\begin{cases} a+c=2\\ d+ac+b=3\\ ad+bc=2\\ bd=-3 \end{cases}

EDIT:

Once this answer is getting attention I will explain more.

Like already said, that is one approach when we don't have any good idea about how to proceed. Solve the system depends how it looks and each system is a new system. However that idea works for many cases that we usually face with.

The first idea to solve the system is look for integer solutions. On that way, the best equation to start is $bd=-3$. That give us the possibilities $(b,d)=\{(-1,3),(1,-3),(-3,1),(3,-1)\}$. Now we plug those possibilities in the system and check if we find $a$ and $c$ that solve it. After do that we see that the solutions are $(a,b,c,d)=\{(1,-1,1,3),(1,3,1,-1)\}$. But both give us the same factorization.

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    $\begingroup$ Simple and brilliant solution. $\endgroup$ – Isai Jan 3 '17 at 20:45
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    $\begingroup$ This is the best solution, since it is the most general method. It should be the accepted one. Keep in mind that solving the set of equations will take some time if done by hand. $\endgroup$ – FundThmCalculus Jan 3 '17 at 21:44
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    $\begingroup$ @FundThmCalculus: Fure sure it cans take a while, but fortunately in many cases is enough to search for integer solutions. It makes the system much easier. $\endgroup$ – Arnaldo Jan 4 '17 at 0:14
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    $\begingroup$ I rather disagree that this is a reasonable solution. Try solving the equations and you will find you end up with a quartic. So in general you've not helped yourself one bit. $\endgroup$ – DRF Jan 4 '17 at 12:09
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    $\begingroup$ @DRF: I totally disagree. I have used this method a bunch of times and it works. Just see my comment above. $\endgroup$ – Arnaldo Jan 4 '17 at 12:13
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Rational Root Theorem Solution

We can reduce the problem to a straightforward Rational Root Theorem problem as follows. We start with reducing the polynomial modulo $x^2 - p x - q$ where $p$ and $q$ are undetermined integers that we want to choose such that this yields zero. This amounts to putting $x^2 = p x + q$ and higher powers are reduced by multiplying this rule by $x$ and then applying this reduction rule again. So, the polynomial becomes a linear function by applying the substitutions:

$$\begin{split} x^2 &= p x + q\\ x^3 &= p x^2 + q x&= (p^2 + q)x + pq\\ x^4 &= p x^3 + q x^2 &= p(p^2+2q)x+q(p^2+q) \end{split}$$

The polynomial reduced modulo $x^2 - p x - q$ then becomes:

$$\left(p^3 + 2 pq + 2 p^2 +2 q+3p + 2\right)x +p^2 q + q^2 + 2 pq + 2 q-3\tag{1}$$

We need to find integers $p$ and $q$ such that the above function is identical zero. We can simplify things by using that the function is then also zero for special values for $x$, allowing us to choose a value for $x$ that simplifies things a lot. Choosing $x = -\frac{q}{p}$ eliminates the $p^3$ term and a few more terms, setting what remains equal to zero yields a linear equation for $p$ in terms of $q$. Expressing $p$ in terms of $q$ yields:

$$p = -\frac{2q(q+1)}{q^2+q+3}\tag{2}$$

We can then substitute this in the constant term of eq. (1), setting the resulting expression equal to zero yields the equation:

$$q^6 + 3 q^5 + 7 q^4 + 10 q^3 - 21 q^2 + 27 q - 27 = 0\tag{3}$$

Here we have set the numerator of the resulting rational function equal to zero. Now we know that $q$ must be an integer, therefore the Rational Root Theorem is guaranteed to succeed and inserting any solution for $q$ (for which denominator is not zero) in eq. (2) will always yield a $p$ such that $x^2 - p x - q$ is a factor of the polynomial. It can be easily shown that this method works for general fourth degree polynomials.

Note also that applying the rational root theorem to find integer solution can be made very efficient by shifting the variable. If the value of a polynomial for some value $u$ is not zero, then this is the value of the constant term of the polynomial obtained by translating the variable by $u$. So, by factoring the value you just obtained you obtain the root candidates of that translated polynomial, adding $u$ to those values yields the root candidates of the original polynomial. The root candidates must then be on the intersection of the original list and the new list. By repeating this method, the list of root candidates will shrink rapidly. This is why I find the Rational Root Theorem such a powerful tool and that's why at least in my opinion it's worthwhile to reduce problems to Rational Root Theorem problems.

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Splitting up the x^2 term works well for factorising quartic but one has to be smart in grouping terms In this , write 3x^2 as x^2 + 2x^2 and group terms as = x^2(x^2+2x+1) +2x^2+2x-3 =[x(x+1)]^2 +2x(x+1) -3

= [x(x+1) +3][x(x+1)-1]

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  • $\begingroup$ Please, use MathJax (i.e. LaTeX commands) for mathematical notations. $\endgroup$ – Taroccoesbrocco May 12 '18 at 8:38
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A general quartic polynomial $ax^4+bx^3+cx^2+dx+e$ can be reduced to the "depressed" form

$$x^4+px^2+qx+r$$ by dividing by $a$ and translating the unknown by $\dfrac b{4a}$.

Now we try the factorization in two quadratic binomials such that the cubic term is missing,

$$(x^2+ux+v)(x^2-ux+w)=x^4+(-u^2+w+v)x^2+u(w-v)x+wv.$$

By identification of the coefficients we obtain

$$\begin{cases}w+v=p+u^2,\\u(w-v)=q\end{cases}$$ and we eliminate $v,w$ with

$$u^2(w+v)^2-u^2(w-v)^2=4wvu^2,$$ i.e.

$$u^2(p+u^2)^2-q^2=4ru^2$$ which is cubic in $u^2$. As the cubic coefficient is positive and the value at $u^2=0$ is negative, there is certainly a positive solution, which we can find by means of the Tartaglia formula.

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