Real root of $f(x) = 1+2x+3x^2 +4x^3$

Question

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.

Answer 1

You're off to a good start. You've proven that there's only one real root as $f'(x)$ is always positive. The next step is to narrow down the interval containing the root. One way to do this is to apply the intermediate value theorem, and find two points where $f(x)$ has different signs.

Observe that $f(0) = 1 > 0$ and $f(-1) = -2 < 0$. This means the root has to be in $(-1, 0)$

Can you narrow it down further?

EDIT: I misread the question to mean asking for the location of the root instead of the sum of real roots. However the answer still works, as there's only one real root.

Answer 2

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