Question: Let $a,b,c$ be real numbers. Such that $a<3$ and all the zeroes of the polynomial $x^3+ax^2+bx+c$ are negative real numbers, then value of $b+c$ can be?
My Attempt: I thought about differentiating $f(x)$ and getting $3x^2+2ax+b$ and had to notice that $f(x)$ should contain one local minima and one local maxima. but I don't know how to relate it with the problem.
Given answer is $1,2 $ and $3$. How do i do it? Please guide.
On
Try writing $$ f(x) = A (x-u) (x- v)(x-w) $$ (in the case where it has three real roots); notice that $A = 1,$ because the leading coefficient of $f$ is $1$, and then that $a = u + v + w$ and $b = uv + vw + uw$ and $c = uvw$. Now $a = 3$ tells you that $$ u+v+w = 3 $$ and you're looking at $$ b+c = uv + vw + uw + uvw $$ Use the first equation to eliminate one variable from the second, and see what you come up with.
(BTW, I don't think that the answer is correct UNLESS the problem says that the roots are negative INTEGERS rather than negative reals...but I haven't actually checked to confirm that).
Let $-u$, $-v$ and $-w$ be our roots.
Hence, $u$, $v$ and $w$ are positive numbers such that $u+v+w<3$
and we need to find the range of $uv+uw+vw+uvw$.
Now, prove that $$0<uv+uw+vw+uvw<4,$$
where in the left inequality $uv+uw+vw+uvw\rightarrow0$ for $u=v\rightarrow0^+$
and in the right inequality $uv+uw+vw+uvw\rightarrow4$ for $u=v=w\rightarrow1^-.$