Calculation:
Condition 1
When x > 0
⇒ x × (6x2 + 1) = 5x2
⇒ 6x3 - 5x2 + x = 0
⇒ x(6x2 - 5x + 1) = 0
⇒ x(3x - 1)(2x - 1) = 0
So the value of x is 0, 1/2, and 1/3 (x ≠ 0)
Condition 2
When x ≤ 0
⇒ -x × (6x2 + 1) = 5x2
⇒ 6x3 + 5x2 + x = 0
⇒ x(6x2 + 5x + 1) = 0
⇒ x(2x + 1)(3x + 1) = 0
So the value of x is 0, -1/2, and -1/3
∴ There are 5 different values of x which are 0, -1/2, 1/2, -1/3, and 1/3.