What are roots of $x^{3}+3x^{2}+4x+1$?

Question

There are no divisor of 1 in this polynomial for which would be satisfied $x^{3}+3x^{2}+4x+1=0$. How to find roots here?

Answer 1

Maybe this could help a bit?

http://www.wolframalpha.com/input/?i=Root[x^3%2B3*x^2%2B4*x%2B1%2C+x]

Answer 2

The equation has one real solution, but rather ugly I am afraid.

http://www4f.wolframalpha.com/Calculate/MSP/MSP41323441hc6heb7fh0000004752i2229938d4f0?MSPStoreType=image/gif&s=27&w=286.&h=81.

Answer 3

Using maxima (I didn't know maxima could do tex(%)).

$$x=-{{{{\sqrt{3}\,i}\over{2}}-{{1}\over{2}}}\over{3\,\left({{ \sqrt{31}}\over{2\,3^{{{3}\over{2}}}}}+{{1}\over{2}}\right)^{{{1 }\over{3}}}}}+\left({{\sqrt{31}}\over{2\,3^{{{3}\over{2}}}}}+{{1 }\over{2}}\right)^{{{1}\over{3}}}\,\left(-{{\sqrt{3}\,i}\over{2}}-{{ 1}\over{2}}\right)-1$$

$$ x=\left({{\sqrt{31}}\over{2\,3^{{{3}\over{2}} }}}+{{1}\over{2}}\right)^{{{1}\over{3}}}\,\left({{\sqrt{3}\,i}\over{ 2}}-{{1}\over{2}}\right)-{{-{{\sqrt{3}\,i}\over{2}}-{{1}\over{2}} }\over{3\,\left({{\sqrt{31}}\over{2\,3^{{{3}\over{2}}}}}+{{1}\over{2 }}\right)^{{{1}\over{3}}}}}-1$$

$$ x=\left({{\sqrt{31}}\over{2\,3^{{{3 }\over{2}}}}}+{{1}\over{2}}\right)^{{{1}\over{3}}}-{{1}\over{3\, \left({{\sqrt{31}}\over{2\,3^{{{3}\over{2}}}}}+{{1}\over{2}}\right) ^{{{1}\over{3}}}}}-1$$

Answer 4

Put $x = y - 1$, this yields:

$$y^3 + y - 1 = 0$$

Then compare with the identity:

$$\begin{split} (a+b)^3 \equiv a^3 + 3 a^2 b + 3 b a^2 + b^3\Longrightarrow \\ (a+b)^3 - 3 a b (a+b) - (a^3 + b^3) \equiv 0 \end{split} $$

So, if we can find two numbers $a$ and $b$ such that:

$$-3 a b = 1$$

and

$$a^3 + b^3 = 1$$

then $y = a + b$ will be a solution.

If we put $A = a^3$ and $B = b^3$, then we require that:

$$A B = -\frac{1}{27}$$

$$A + B = 1$$

Now solving for $A$ and $B$ requires solving a quadratic equation. When extracting $a$ and $b$, you must make sure that $a b = -\frac{1}{3}$, so the choice of the phase factor multiplying the cube root for $A$ fixes the phase factor for $b$.

Answer 5

This can be done by hand with a bunch of substitutions.

Let $y=x+1$, then

$$1+4(y-1)+3(y-1)^2+(y-1)^3=0$$

If we expand all the terms we get

$$1+4y-4+3y^2-6y+3+y^3-3y^2+3y-1=0$$

which boils down to

$$y^3+y-1=0$$

Now we can do a change in coordinates with another substitution.

Let $y=u+\frac{\lambda}{u}$. We will find $\lambda$ later. This yields.

$$-1+u+\frac{\lambda}{u}+\left ( u + \frac{\lambda}{u} \right )^3=0$$

We can expand this by multiplying by $u^3$ which gives

$$u^4(3\lambda + 1) + u^2\lambda(3\lambda+1)-u^3+u^6+\lambda^3=0$$

We want the coefficients on the $u^2$ and $u^4$ terms to be $0$. So $\lambda=-\frac{1}{3}$. Substituting we now have

$$u^6-u^3-\frac{1}{27}=0$$

If we perform yet another substitution, we can finally get to a quadratic.

Let $z=u^3$.

$$z^2-z+\frac{1}{27}=0$$

Using the quadratic formula we get

$$z=\frac{1}{18}(9+\sqrt{93})$$.

Now we climb back up our ladder of subsitutions

$$u^3=\frac{1}{18}(9+\sqrt{93})$$

$$u=\sqrt[3]{\frac{1}{18}(9+\sqrt{93})}$$

We substitute for $u$ and get

$$y=\sqrt[3]{\frac{1}{18}(9+\sqrt{93})}-\frac{1}{3\sqrt[3]{\frac{1}{18}(9+\sqrt{93})}}$$

Now to get $x$ we just subtract 1.

$$x=\left( \frac{1}{18}(9+\sqrt{93})\right )^{1/3} - \frac{1}{3}\left( \frac{1}{18}(9+\sqrt{93})\right )^{-1/3}-1$$