Let r1, r2 and r3 be the solutions of the equation x3-2 x2+4 x+5074=0 then the value of (r1+2)(r2+2)(r3+2) is

Question

Let r1, r2 and r3 be the solutions of the equation x3-2 x2+4 x+5074=0 then the value of (r1+2)(r2+2)(r3+2) is (A) 5050 (B) 5066 (C) -5050 (D) -5066

Tardigrade Admin · Accepted Answer

x3-2 x2+4 x+5074=(x-r1)(x-r2)(x-r3) text put x=-2 -8-8-8+5074=-(2+r1)(2+r2)(2+r3) ∴ 5050=-(2+r1)(2+r2)(2+r3) (2+r1)(2+r2)(2+r3)=-5050

Q. 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 $(r_{1} + 2) (r_{2} + 2) (r_{3} + 2)$ is