Could someone please explain how to solve this : $x^4+3x^3-6x^2+16x+56=0$ – not the answer only, but a step-by-step solution.
[Math] Solving Quartic Equation
algebra-precalculuselementary-number-theorypolynomials
Related Solutions
Since the polynomial has integer coefficients, the rational root theorem applies. Thus any rational root must be of the form $x=\pm p/q$, where $p$ divides the constant term 100 and $q$ divides the leading coefficient 1. In this case, the only possibility for $q$ is 1. This tells you that any rational root must be a divisor of $100=2^2*5^2$. It turns out that this polynomial does have rational roots, after which you find one you can perform polynomial division to get a complete factorization.
For instance, we have the potential rational roots $x=\pm2,\pm5,\pm10,\pm20\pm25,\pm50,\pm100$. We could plug in $x=5$ and verify that this is a root. Then, $$ \frac{x^4 - 10x^3 + 21x^2 + 40x - 100}{x-5} = x^3-5x^2-4x+20. $$ Since all the roots are rational, repeating this process will generate all of them. Not every polynomial with integer coefficients has rational roots (for instance $x^2-2=0$), so this won't always be the case.
$$e^{\frac{x^2}{4vt}} = 1+\frac{x^2}{2vt}$$ Let $y=\frac{x^2}{4vt}$ $$e^y=1+2y$$ $$e^{-y}=\frac{1}{1+2y}$$ $$(1+2y)e^{-y}=1$$ $$(\frac12+y)e^{-y}=\frac12$$ $$(-\frac12-y)e^{-y}=-\frac12$$ $$ (-\frac12-y)e^{-y}e^{-\frac12}=-\frac12 e^{-\frac12}$$ $$ (-\frac12-y)e^{-\frac12-y}=-\frac12 e^{-\frac12}=-\frac{1}{2\sqrt{e}}$$ $X=(-\frac12-y)$ $$Xe^X=-\frac{1}{2\sqrt{e}}$$ From the definition of the Lambert W function : http://mathworld.wolfram.com/LambertW-Function.html $$X=W\left(-\frac{1}{2\sqrt{e}}\right)$$ $$y=-\frac12-X=-\frac12-W\left(-\frac{1}{2\sqrt{e}}\right)$$ $$x=\sqrt{4vty}=\sqrt{-2-4W\left(-\frac{1}{2\sqrt{e}}\right)}\sqrt{vt}$$ The Lambert W(z) function is a multi valuated function in $-\frac{1}{e} <z<0$ , real $z$.
This is presently the case where $z=-\frac{1}{2\sqrt{e}}$ since $-\frac{1}{e} <-\frac{1}{2\sqrt{e}}<0$
First root :
$W_0\left(-\frac{1}{2\sqrt{e}}\right)=-\frac12 \quad;\quad {-2-4W_0\left(-\frac{1}{2\sqrt{e}}\right)}=-2-4(-1/2)=0 \quad;\quad x=0$
Second root :
$W_{-1}\left(-\frac{1}{2\sqrt{e}}\right)\simeq -1.756431... $
https://www.wolframalpha.com/input/?i=lambertw(-1,-1%2F(2+sqrt(e)))
One can use series expansion to compute it approximately. See page 13 in https://fr.scribd.com/doc/34977341/Sophomore-s-Dream-Function . The convergence is very slow. See the numerical calculus in the below addition. In practice, it is certainly faster to solve directly $e^y=1+2y$ with Newton-Raphson or similar iterative method.
Finally, an approximate value is :
$$x\simeq\sqrt{-2-4(-1.756431)}\sqrt{vt}\simeq 2.2418128 \sqrt{vt}$$
With more digits : https://www.wolframalpha.com/input/?i=sqrt(-2-4+lambertw(-1,-1%2F(2+sqrt(e))))
IN ADDITION :
Example of recursive numerical calculus of $W_{-1}(x)$ in the range $-\frac{1}{e}<x<0$
Best Answer
All quartic equations can be solved with radicals,
See the following:
If you don't want to follow that route I suggest the following. Put the polynomial in a calculator and look at the x-intercepts. If this is a homework problem one of those will be whole number or fraction. Then use long division to factor the polynomial further.
As an example if you plugged $9x^2-1=0$ into a calculator you would see it had an x-intercept at $x=0.33333333 \approx \frac{1}{3}$. Plugging $x=1/3$ into the equation would yield $0=0$ as expected. This tells us to divide by $x-\frac{1}{3}$, doing this will give a qoutient of $x+\frac{1}{3}$ which tells us the other root.
See: wolframalpha for the actual roots which are disgustingly complicated.
I decided to add a example of using Ferrari's method to solve the quartic. However I will be solving a different quartic from the one posted.
Consider the equation below,
$$ x^4+x^3+x^2+x+1=0$$
Our first step is to make the substitution $x=t-\frac{b}{4a}$ where b=1 and a=1.
$$ (t-\frac{1}{4})^4+(t-\frac{1}{4})^3+(t-\frac{1}{4})^2+(t-\frac{1}{4})+1=0$$
$$ t^4 + \frac{5}{8} t^2 +\frac{5}{8}t+\frac{205}{256} = 0 \qquad \text{(notice we lost } t^3)$$
Now push the linear term to the right hand side and complete the square on the left.
$$ t^4 + \frac{5}{8} t^2 +\frac{205}{256} = -\frac{5}{8}t$$
$$ (t^2+\frac{5}{16})^2 + 45/64 = -\frac{5}{8}t$$
$$ (t^2+\frac{5}{16})^2 = -\frac{5}{8}t - \frac{45}{64}$$
We now add and subtract an as yet unknown variable $z$ within the squared term.
$$ (t^2+\frac{5}{16}+z-z)^2 = -\frac{5}{8}t - \frac{45}{64}$$
Note that $(t^2+\frac{5}{16}+z-z)^2 = (t^2+\frac{5}{16}+z)^2 -2z(t^2+\frac{5}{16}+z)+z^2 = (t^2+\frac{5}{16}+z)^2 - 2z(t^2+\frac{5}{16})-z^2$
Putting this into our expression and isolating the perfect square we get,
$$ (t^2+\frac{5}{16}+z)^2 = -\frac{5}{8}t - \frac{45}{64}+2z(t^2+\frac{5}{16})+z^2$$
The left is a perfect square in the variable $t$. This motivates us to rewrite the right hand side in that form as well. Therefore we require that the discriminant $B^2-4AC$ be zero.
$$ A=2z, B=-\frac{5}{8}, C=z^2+\frac{5}{8}z-\frac{45}{64}$$
$$ \frac{25}{64} - 8z(z^2+\frac{5}{8}z-\frac{45}{64}) = 0 $$
$$ -8 z^3 -5 z^2 + \frac{45}{8} z + 25/64 = 0 $$
We need to find the $z$ which satisfy this equation. There is another formula (called cardono's formula) which can solve for these). Fortunately this polynomial has a rational root of $z=5/8$.
We will put z=5/8 back into our original equation and get,
$$ (t^2+\frac{15}{16})^2 = 5/64-(5 t)/8+(5 t^2)/4$$
$$ (t^2+\frac{15}{16})^2 = \frac{5}{4}(t-\frac{1}{4})^2$$
Taking the square root of both sides we get,
$$ t^2+\frac{15}{16} = \pm\frac{\sqrt{5}}{2}t \mp \frac{\sqrt{5}}{8}$$
$$ t^2 \mp \frac{\sqrt{5}}{2}t +\frac{15}{16} \pm \frac{\sqrt{5}}{8} = 0$$
And now applying the good old quadratic formula we get,
$$t = \frac{\pm\frac{\sqrt{5}}{2} \pm \sqrt{\frac{5}{4}-4(\frac{15}{16} \pm \frac{\sqrt{5}}{8})} }{2}$$
One of these is, $$ t = \frac{\sqrt{5}}{4}+i \frac{\sqrt{5+\sqrt{5}}}{2\sqrt{2}} = e^{2\pi i/5} + \frac{1}{4}$$,
Recall that our original variable was $x=t-\frac{1}{4}$.
A solution to our original quartic is $x=e^{2 \pi i /5}$. Substituting this in the original gives,
$$ e^{8\pi i/5}+e^{6\pi i/5} + e^{4\pi i/5} + e^{2\pi i/5} + 1 = 0 $$
Which can be verified at wolframalpha.
I pretty much just followed the instructions in my first link, but I thought it might be helpful to have another example.