Question

Consider the polynomial $f\left(x\right)=1+2x+3x^2+4x^3.$ Let $s$ be the sum of all distinct real roots of $f\left(x\right)=0$ and let $\left|s\right|=t.$

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 1

Answer: (C)

$(-\frac{3}{4},-\frac{1}{2})$

Given, $f\left(x\right)=4x^3+3x^2+2x+1\Rightarrow f^{\prime }\left(x\right)=2(6x^2+3x+1)$
$\Rightarrow D=9-24<0$
Hence, $f\left(x\right)=0$ has only one real root
$f(-\frac{1}{2})=1-1+\frac{3}{4}-\frac{4}{8}>0$
$f(-\frac{3}{4})=1-\frac{6}{4}+\frac{27}{16}-\frac{108}{64}=\frac{64-96+108-108}{64}<0$
$f\left(x\right)$ changes its sign in $(-\frac{3}{4},-\frac{1}{2})$
Hence $f\left(x\right)=0$ has a root in $(-\frac{3}{4},-\frac{1}{2})$