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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?

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    Hint: can a real polynomial have one single complex non-real root? – 2017-01-28
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    Complex roots occur in pairs. Slipped my mind. Thanks – 2017-01-28
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    Right. Now think what happens for very large $x$, positive or negative, and you'll figure out where the third real root is. – 2017-01-28
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    @dxiv By my calculations, for reasonably sized $a,b,c$ such that their absolute values are at most $11!!!!!!!!!$, then the root shall lie between $x=-1$ and $x=-\text{Graham's number}$ :-) – 2017-01-28
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    @SimplyBeautifulArt That's always true, for suitable values of Graham's number ;-) – 2017-01-28

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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.