How to solve exercises with polynomial with 2 parameters having all real roots? $f=2x^{4}+4x^{3}+3x^{2}+bx+c$ has all real roots, find b,c (b,c are from R).
Thanks a lot, I tried with substitution, I don't know, is there something with derivative? please help thanks
 A: Let $g(x)=2x^4+4x^3+3x^2$.  Then $g''(z)=24x^2+24x+6=6(2x+1)^2\geq0$.  So $g$ is convex, and will intersect a line at most twice.  In order for $f(x)$ to have only real roots, the line $y=-bx-c$ will have to be tangent to the graph of $g$.  But the tangent will not intersect the graph in a second point, so if $f$ has $4$ real zeros, it must have a quadruple zero, that is $f$ and its first three derivatives must vanish at some point.
The computation above shows that this point must be $x=-\frac12$ and we must have $$f(x)=2\left(x+\frac12\right)^4.$$
I haven't carried it past this point.  I leave it to you.  
A: $f$ has $4$ (not necessarily distinct) real roots, say $r, s, t, u$. Hence, $f$ can be factorized this way : $f = 2 (x-r) (x -s) (x-t) (x - u)$. If you developp this expression, and compare term by term, you'll get that the roots satisfy he following system of equations (edited 4 equations, as pointed out by @saulspatz) :
$$\left\{ \begin{array}{ccl} 
c & =  & 2 rstu \\
b & = & -2(rst + rsu + rtu + stu) \\
3 & = & 2 (rs + rt + ru + st + su + tu) \\
4 & = & -2(r + s + t + u) \end{array}\right. $$
Granted, this looks ugly, but only the last two lines represents a constraint. e.g. take $r = s = t = u = \frac{-1}{2}$, this satisfies the two last lines and hence you get a solution where $b = 1$ and $c = \frac{1}{8}$. 
A: $f(x) = 2x^4+4x^3+3x^2+bx+c$
If $f(x)$ has all $4$ roots to be real and complex, then it's discriminate $$\Delta = (-72)b^2+176b^3-108b^4-864c-2304bc+1536b^2c+2304c^2-3072bc^2+2048c^3$$ must be real, $\Delta > 0$
Also if you divide $f(x)$ by 2, and make a translation $x = y-\frac{1}{2}$
$2x^4+4x^3+3x^2+bx+c = 0$
$x^4+2x^3+\frac{3/2}x^2+\frac{b/2}x+\frac{c/2} = 0$
Say $x = y-\frac{1}{2}$
The equation is depressed that suddenly appear to be
$y^4+(\frac{b}{2}-\frac{1}{2})y+\frac{c}{2}-\frac{b}{4}+\frac{3}{16} = 0$
Notice that the $y^2$ automatically vanishes
Defined $D = 512c-256b+192$
For the polynomial to therefore have real distinct root $D < 0$ and $\Delta > 0$ 
Nature of roots
