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So I was playing around with solving polynomials last night and realized that I had no idea how to solve a polynomial with no rational roots, such as $$x^4+3x^3+6x+4=0$$ Using the rational roots test, the possible roots are $\pm1, \pm2, \pm4$, but none of these work.

Because there were no rational linear factors, I had to assume that the quartic separated into two quadratic equations yielding either imaginary or irrational "pairs" of roots. My initial attempt was to "solve for the coefficients of these factors".

I assumed that $x^4+3x^3+6x+4=0$ factored into something that looked like this $$\left(x^2+ax+b\right)\left(x^2+cx+d\right)=0$$ because the coefficient of the first term is one. Expanding this out I got $$x^4+ax^3+cx^3+bx^2+acx^2+dx^2+adx+bcx+bd=0$$ $$x^4+\left(a+c\right)x^3+\left(b+ac+d\right)x^2+\left(ad+bc\right)x+bd=0$$ Equating the coefficients of both equations $$a+c = 3$$ $$b+ac+d = 0$$ $$ad+bc = 6$$ $$bd = 4$$ I found these relationships between the various coefficients. Solving this system using the two middle equations: $$\begin{cases} b+a\left(3-a\right)+\frac4b=0 \\ a\frac4b+b\left(3-a\right)=6 \end{cases}$$ From the first equation: $$a = \frac{3\pm\sqrt{9+4b+\frac{16}{b}}}{2}$$ Substituting this into the second equation: $$\frac{3\pm\sqrt{9+4b+\frac{16}{b}}}{2}\cdot\frac4b+b\cdot\left(3-\frac{3\pm\sqrt{9+4b+\frac{16}{b}}}{2}\right)=6$$ $$3\left(b-2\right)^2 = \left(b^2-4\right)\cdot\pm\sqrt{9+4b+\frac{16}b}$$ $$0 = \left(b-2\right)^2\cdot\left(\left(b+2\right)^2\left(9+4b+\frac{16}b\right)-9\left(b-2\right)^2\right)$$ So $b = 2$ because everything after $\left(b-2\right)^2$ did not really matter in this case. From there it was easy to get that $d = 2$, $a = -1$ and $c = 4$. This meant that $$x^4+3x^3+6x+4=0 \to \left(x^2-x+2\right)\left(x^2+4x+2\right)=0$$ $$x = \frac12\pm\frac{\sqrt7}{2}i,\space x = -2\pm\sqrt2$$

These answers worked! I was pretty happy at the end that I had solved the equation which had taken a lot of work, but my question was if there was a better way to solve this?

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    $\begingroup$ The coefficient of the first term is one, right? $\endgroup$
    – Pedro
    Commented Dec 31, 2014 at 22:54
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    $\begingroup$ Yes it is suppose to be one :) The little things are always easy to miss. $\endgroup$
    – dardeshna
    Commented Dec 31, 2014 at 22:56
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    $\begingroup$ has no positive roots and all real roots are in $(-5, 0)$ $\endgroup$
    – abel
    Commented Dec 31, 2014 at 23:24
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    $\begingroup$ You could use the general solution to the quartic, but few people would call that easier. $\endgroup$ Commented Jan 1, 2015 at 0:06
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    $\begingroup$ The next reasonable guess after looking for rational roots would be to suppose that at least the coefficients are integers. So after the system of equations ending in $bd=4$, I would start guessing $(b,d) = (4,1)$, $(b,d) = (2,2)$, etc., and indeed, this second guess is right. $\endgroup$
    – Théophile
    Commented Nov 11, 2015 at 23:09

4 Answers 4

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Hint:

First check that $0$ is not a solution, hence $x\neq0\,$, so it is legal to divide by $x^2$. We get

$$ x^2+3x+\frac6x+\frac4{x^2}=0. \tag1 $$

Now note that

$$ \left(x+\frac2x\right)^2=x^2+4+\dfrac{4}{x^2}\iff x^2+\dfrac{4}{x^2}=\left(x+\dfrac2x\right)^2-4. $$

So $(1)$ can be written as :

$$ \left(x^2+\dfrac4{x^2}\right)+\left(3x+\dfrac6x\right)=0\iff \left(x+\dfrac2x\right)^2-4+3\left(x+\dfrac2x\right)=0. $$

Now use the substitution $u=x+\frac2x$ and you get the quadratic :

$$ u^2+3u-4=0. $$

Answer to a comment:

Awesome, but how did you see that? And is this just then a specific case...?

I remarked that the coefficients of the equation were symmetric in the following sense :

$$ x^4+3x^3+6x+4=0\iff (x^4+\color{#C00}2^2)+3(x^3+\color{#C00}2x)=0. $$

So I tried to divide by $x^2$, then to find a relation between $x^2+\tfrac4{x^2}$ and $\left(x+\tfrac2x\right)^2$, so that I can convert it into a quadratic.

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    $\begingroup$ The algebra worked for me. Great answer! $\endgroup$
    – graydad
    Commented Dec 31, 2014 at 23:07
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    $\begingroup$ If $x=0$ is a solution then the last term would not be $4$ so there is not very much to check. $\endgroup$ Commented Dec 31, 2014 at 23:16
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    $\begingroup$ Awesome, but how did you see that? And is this just then a specific case...? $\endgroup$
    – dardeshna
    Commented Dec 31, 2014 at 23:47
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    $\begingroup$ @dardeshna I explained that after my recent edit. $\endgroup$
    – Workaholic
    Commented Jan 1, 2015 at 12:18
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Reorganize the term like $$(x^4+4+4x^2)+(3x^3+6x)-4x^2=(x^2+2)^2+3x(x^2+2)-4x^2$$ than it is easy to come up with $$(x^2+4x+2)(x^2-x+2)$$

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In general any quartic equation can be solved. If the polynomial happens to be the product of two irreducible polynomials over $\mathbb Z$, the following method allows to find the coefficients easily. Reducing the coefficients mod $2$, one has $$f(x):=x^4+3x^3+6x+4=x^4+x^3=x^2(x^2+x).$$ Lifting back to ${\mathbb Z}$-coefficients, one may assume that $$f(x)=(x^2+2ax\pm 2)(x^2+bx\pm 2),$$ where $a,b\in {\mathbb Z}$ and $b$ is odd. By comparison of coefficients, one has $$2a+b=3,2ab\pm 4=0,\pm 4a\pm 2b=6,$$ which shows immediately that one needs to take the ‘plus’ sign from $\pm$. From the first two equations, by eliminating $a$, one has $b=4$ or $b=-1$, so $b=-1$ since it is odd. It follows that $a=2,b=-1$ and $$f(x)=(x^2+4x+2)(x^2-x+2).$$

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More generally:

Given a quartic equation with the form

$x^4+bx^3+cx^2+(br)x+r^2=0$

you guarantee factors with the form

$(x^2+px+r)(x^2+qx+r)=0$

For then the cubic and linear terms are both matched by rendering

$p+q=b$

and we have only to match the quadratic terms which oeadsto

$pq=c-2r^2$

From these equations $p$ and $q$ are roots of the quadratic equation

$u^2-bu+(c-2r^2)=0.$

Here $b=3,c=0,r=2$, from which

$u^2-3u-4=0.$

Thus $\{p,q\}=\{-1,4\}$ from which we obtain the factorization

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

The quartic equation is then solved by the roots of these quadratic factors.

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