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.
Best Answer
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.