When 6x^3-7x^2+1=0 whats the value of (2x+1)/x^2

i have choices. A)1 B)2 C)3 D)4 E)5

i also found that 6x^3-7x^2+1 polynomial equals to (3x+1)*(2x-1)*(x-1)

When i use -1/3 and 1 for x i get 3 in the second division but when i use 1/2 for x i get 8. How can i solve this problem?

6x³-7x²+1=0
6=1*2*3
x=±p/q, p=1 and q=1*2*3

Let x=1
6*1³-7*1²+1=0 hence x₁=1 is root.

(6x³-7x²+1)/(x-1)=6x²-x-1
6x²-x-1=0
D=(-1)²-4*6*(-1)=5²
x₂=(1-5)/12=-1/3
x₃=(1+5)/12=½

hence
(6x³-7x²+1)=6(x-1)(x-½)(x+1/3) ok!

let x=1
(2x+1)/x²=(2*1+1)/1²=3

let x=½
(2x+1)/x²=(2*(½)+1)/(½)²=8

let x=-1/3
(2x+1)/x²=(2*(-1/3)+1)/(-1/3)²=3

ok!

Set C)3