Q. $F(x)= x^3+ ax^2 +bx+ c$ where $a$, $b$ and $c$ are real numbers. Suppose $c<0$, $a+b+c>-1$ and $a-b+c>1$ then what can we conclude about its roots.
Ans. I know that $f(-1)>0$, $f(0)<0$ and $f(1)>0$ so it has two real root one positive and other negative. But what about the third root is it real or imaginary?
There are three roots.
If the third root is not real, but the other two roots are real, it follows that the sum of the roots is not real. However, $-a$ is a real number, so this is a contradiction.
We can also ascertain that there are three roots. If it has two roots, than $f(x)$ is of the form $$f(x)=(x-\alpha)^2(x-\beta)$$ Then $f(0)=-\alpha^2 \beta<0$, $f(1)=(1-\alpha)^2(1-\beta)>0$, $f(-1)=(-1-\alpha)^2(-1-\beta)>0$. So we have $$\beta>0, 1>\beta , -1>\beta$$ Which is a contradiction.
