Solve $x^4+3x^3+6x+4=0$… easier way?

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 $$(x^2+ax+b)(x^2+cx+d)=0$$ because the coefficient of the first term is one. Expanding this out I got $$x^4+ax^3+cx^3+bx^2+acx^2+dx^2+adx+bcx+bd=0$$ $$x^4+(a+c)x^3+(b+ac+d)x^2+(ad+bc)x+bd=0$$ 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: $$\begin{cases} b+a(3-a)+\frac4b=0 \\ a\frac4b+b(3-a)=6 \end{cases}$$ From the first equation: $$a = \frac{3\pm\sqrt{9+4b+\frac{16}{b}}}{2}$$ Substituting this into the second equation: $$\frac{3\pm\sqrt{9+4b+\frac{16}{b}}}{2}\cdot\frac4b+b\cdot(3-\frac{3\pm\sqrt{9+4b+\frac{16}{b}}}{2})=6$$ $$3(b-2)^2 = (b^2-4)\cdot\pm\sqrt{9+4b+\frac{16}b}$$ $$0 = (b-2)^2\cdot((b+2)^2(9+4b+\frac{16}b)-9(b-2)^2)$$ So $b = 2$ because everything after $(b-2)^2$ 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 $$x^4+3x^3+6x+4=0 \to (x^2-x+2)(x^2+4x+2)=0$$ $$x = \frac12\pm\frac{\sqrt7}{2}i,\space x = -2\pm\sqrt2$$

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?

• The coefficient of the first term is one, right? – Pedro Dec 31 '14 at 22:54
• Yes it is suppose to be one :) The little things are always easy to miss. – dardeshna Dec 31 '14 at 22:56
• has no positive roots and all real roots are in $(-5, 0)$ – abel Dec 31 '14 at 23:24
• You could use the general solution to the quartic, but few people would call that easier. – Rory Daulton Jan 1 '15 at 0:06
• 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. – Théophile Nov 11 '15 at 23:09

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.

• The algebra worked for me. Great answer! – graydad Dec 31 '14 at 23:07
• If $x=0$ is a solution then the last term would not be $4$ so there is not very much to check. – Suzu Hirose Dec 31 '14 at 23:16
• Awesome, but how did you see that? And is this just then a specific case...? – dardeshna Dec 31 '14 at 23:47
• @dardeshna I explained that after my recent edit. – Workaholic Jan 1 '15 at 12:18

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 $$(x^2+4x+2)(x^2-x+2)$$