I have not been able to crack this question for a long time. Initially, I thought of using Remainder theorem in the question, but after some time I realized that my thought process for the question was inconclusive.
Let $f(x)$ be a cubic polynomial $x^3+ax^2+bx+c$ such that $f(x)=0$ has three distinct integral roots an $f(g(x)) = 0$ has no real roots, where $g(x)=x^2+2x-5$, then what is the minimum value of $a+b+c$. Can you please tell me about any approach with which I can start the question?
Let the roots of $f$ be $r, s, t.$ We know that $x^2 + 2 x - 5 \neq r, s, t,$ for any real $x,$ which measn that the discriminant, which is equal to $6 +(r, s, t)$ is always negative. This means that the biggest root of $f$ is at least $-7.$
The sum of $a, b, c$ is the value of the polynomial at $1$, less $1,$ so $(1-r)(1-s)(1-t)-1.$ Clearly, the value is minimal when $r=-7, s=-8, t=-9,$ and is equal to $719.$
Hints:
express $a,b,c$ in function of $p,q,r$, the roots
if $f(g(x))=0$ has 0 solutions, $p,q,r$ can't be in $Im(g)$.
minimize $a+b+c$ with previous item