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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, 2017 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, 2017 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, 2017 at 17:15
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    $\begingroup$ If you know the roots, then you can also factorize it. $\endgroup$
    – YukiJ
    Jan 3, 2017 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, 2017 at 19:22

5 Answers 5

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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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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, 2017 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. $\endgroup$ Jan 3, 2017 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, 2017 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, 2017 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, 2017 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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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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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. $\endgroup$ May 12, 2018 at 8:38
  • $\begingroup$ In this particular problem, the suppressed quartic equation reduces to a bi-quadratic equation having no "u" term. It has only u^4 , u^2 and constant term. We can easily solve this equation by quadratic formula and and translate it back to the original variable. The equation has two conjugate irrational and two conjugate complex roots. $\endgroup$ Jun 8, 2021 at 16:00

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