Let $f(x) = x^3+ax^2+bx+c$ where $a,b$ and $c$ are real numbers. Suppose $c<0, a+b+c>-1$ and $ a-b+c>1.$ Then which of the following are true?

Question

(a) All roots of $f(x)$ are real.
(b) $f(x)$ has one real root and $2$ complex roots.
(c) $f(x)$ has two roots in $(-1,1).$
(d) $f(x)$ has at least one negative root.

I thought of solving this question using Descartes Rule. 'c' is negative and 'a' is turning out to be positive. 'b' should be negative but I'm not sure about my work.
Also any literature or links regarding this concept will be appreciated.

Answer 1

Compare

$f(0) = c <0$

$f(1) = 1 + a + b+ c > 1 + -1 > 0$

$f(-1) = -1 +a - b + c > -1 + 1 > 0$.

So there is a root in $(-1,0)$ and in $(0, 1)$.

So that tells us b) is false, d) is true

Complex roots come in pairs and there are at most three roots so there are no complex roots.

So that gives us a) is true.

As $x \to -\infty$ we have $x^3 + ax^2 +bx + c \to -\infty$ and $f(-1) >0$ that means there is a third real root in $(-\infty, -1)$.

So that tells as c) is true.

So we know there are exactly three real roots, two negative-- one less than -1; one in $(-1,0)$, and one positive in $(0,1)$.

So we didn't need descartes rule of signs after all but if we had tried it:

You can use descartes rules of sign that $(a+b+c) + (a-b+c) = 2(a+c) > 0$ so $2a > - c > 0$ so $a > 0$. We can not determine what be is but $b$ is eithere $0$ or the same sign as $c$ or the same sign as $a$. So it doesn't make a sign change.

If $(+x)$ there is one sign so at most one real positive root. $(-x)$ has two sign changes and so at most two real negative roots.

Answer 2

So $f(1) >0$ and $f(-1)>0$ and $f(0)<0$:

\begin{array}{cccc} x & -1 & 0 & 1 \\ f(x) & + & - & + \\ \end{array}

So $a)$ and $c)$ and $d)$ are true.