How do I factor a polynomial over the reals? $$2x^4-5x^3-4x^2+15x-6$$
I got that there are 3 positive roots, and 1 negative root.
I also got that the possible rational roots are $$\pm\frac{1}{2},\pm1,\pm2,\pm\frac{3}{2},\pm3,\pm6$$
Am I just supposed to plug everything in?
 A: $2x^{4}-5x^{3}-4x^{2}+15x-6=2(x^{4}-2x^{2}-3)-5x(x^{2}-3)=2(x^{2}-3)(x^{2}+1)-5x(x^{2}-3)$
$=(x^{2}-3)(2(x^{2}+1)-5x)=(x-\sqrt{3})(x+\sqrt{3})(2x^{2}-5x+2)=(x-\sqrt{3})(x+\sqrt{3})(2x-1)(x-2)$
A: Sometimes, you'll "see" something clever you can do, as did user71352 found. That can save one a good amount of work. 
However, when inspiration has escaped us, one more methodological route to go with this type of a problem is to "test out" possible rational roots $x_i$ of your polynomial $p(x) = 2x^4-5x^3-4x^2+15x-6$, by substituting "candidates" into $p(x)$ and testing if $p(x_i) = 0.\;$ You've done a nice job of determining the candidates. Testing promising roots, you'll find that $x_1 = \frac 12$ and $x_2 = 2$ are both roots. 
Once you have even just one root, say $x = a$, divide a given polynomial by the factor $(x - a)$. In our case, we know two factors, each factor corresponding to one of the two roots we have: $(x - \frac 12)$ and $(x - 2)$. 
So we can divide your polynomial $p(x)$ by the product of the factors  $$\left(x - \frac 12\right)(x - 2) =x^2 - \frac 52 x + 1$$ to determine the remaining quadratic:$$\frac{2x^4-5x^3-4x^2+15x-6}{x^2 - \frac 52 x + 1} = 2x^2 - 6 = 2(x^2 - 3).$$ This is a difference of squares, which factors nicely, thus revealing the roots $x_3,\; x_4 = \pm \sqrt 3,\,$respectively, which are the two remaining roots of $p(x)$.
To summarize, we have the polynomial $$p(x) =  2x^4-5x^3-4x^2+15x-6 = 2\left(x - \frac 12\right)(x -2)(x + \sqrt 3)(x -\sqrt 3)$$ with roots $$x_1 = \frac 12, \;\;x_2 = 2,\;\;x_3 = -\sqrt 3,\;\;x_4 = \sqrt 3$$
