Question

what are the real solutions to
2x^2(x-5)(x+2)

Answer 1

Assuming you mean solve 2x^2(x−5)(x+2) = 0, set each factor to zero:

- 2 ≠ 0
- x^2 = 0 → x = 0 (double root)
- x − 5 = 0 → x = 5
- x + 2 = 0 → x = −2

So the real solutions are x = −2, 0, and 5 (with x = 0 of multiplicity 2).