<s>[INST] Real root of $f(x) = 1+2x+3x^2 +4x^3$ Consider the polynomial $f(x) = 1+2x+3x^2 +4x^3$. Let $s$ be the sum of all distinct real roots of $f(x)$ and let $t = |s|$.
The real number $s$ lies in the interval
A) $\ \ (-\frac{1}{4},0)$ 
B) $ \ \ (−11,−\frac{3}{4}) $
C) $\ \ (−\frac{3}{4},−\frac{1}{2})$
D) $\ \ (0,\frac{1}{4})$
Answer is 'C'
My approach, $f'(x)=2+6x+12x^2$
$D<0$ for $f'(x)$ hence $f(x)$ has one real root. The root of $f(x)$ are $a , c+id$ and  $c-id$. 
$a+2c=-\frac{3}{4}$ and $a(c^2-d^2)=-1$
I am not able to proceed from here.
 [/INST] f(x) is a polynomial function and so it is continuous and differentiable through out. We have by intermediate value theorem, if f is continous in $[a,b]$ and if $f(a)<0, f(b)>0$ then there exists $c \in (a,b)$ such that f(c) = 0
in above options, consider (C):
$f(-1/2) = positve$
$f(-3/4) = negative$
So there should exist by intermediate value theorem, $c \in (-3/4, -1/2)$ such that f(c)=0
</s>