Question

Let $r_1, r_2$ and $r_3$ be the solutions of the equation $x^3-2 x^2+4 x+5074=0$ then the value of $\left(r_1+2\right)\left(r_2+2\right)\left(r_3+2\right)$ is

• A. 5050
• B. 5066
• C. -5050
• D. -5066

Answer 1

Answer: (C)

$x^3-2 x^2+4 x+5074=\left(x-r_1\right)\left(x-r_2\right)\left(x-r_3\right) $
$\text { put } x=-2$
$-8-8-8+5074=-\left(2+r_1\right)\left(2+r_2\right)\left(2+r_3\right) $
$\therefore 5050=-\left(2+r_1\right)\left(2+r_2\right)\left(2+r_3\right)$
$\left(2+r_1\right)\left(2+r_2\right)\left(2+r_3\right)=-5050$