Answer:

Step-by-step explanation:
2x³+ 6x² - x - 10 = 0
(1) Possible roots
The Rational Roots Theorem states that, if a polynomial has any rational roots, they will have the form p/q, where p is a factor of the constant term and q is a factor of the leading coefficient.

In your function, the constant term is -10 and the leading coefficient is 2, so

Factors of 10 = ±1, ±2, ±5, ±10
Factors of 2 = ±1, ±2

(2) Synthetic division
Rather than work through all 12 possibilities, I will do one that works.

So, x = -2 is a root, and the quotient is 2x² + 2x - 5.
(3) Check for other rational roots
2x² + 2x - 5 = 0
D = b² - 4ac =2²- 4(2)(-5) = 4 + 40 = 44
√44 = 2√11, which is irrational.
Since irrational roots come in pairs, the cubic equation has two real, irrational roots and one rational root at x = -2.