Answer:
[tex]\boxed{(1) \, x = \, \pm \dfrac{1}{2}, \pm 1, \pm2, \pm \dfrac{5}{2}, \pm 5, \pm 10; (2) \, x = -2}[/tex]
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.
[tex]\text{Possible rational root} = \dfrac{ p }{ q } = \dfrac{\text{factor of constant term}}{\text{factor of leading coefficient}}[/tex]
In your function, the constant term is -10 and the leading coefficient is 2, so
[tex]\text{Possible root} = \dfrac{\text{factor of 10}}{\text{factor of 2}}[/tex]
Factors of 10 = ±1, ±2, ±5, ±10
Factors of 2 = ±1, ±2
[tex]\text{Possible roots are } \large \boxed{\mathbf{x = \pm \dfrac{1}{2}, \pm 1, \pm2, \pm \dfrac{5}{2}, \pm 5, \pm 10}}[/tex]
(2) Synthetic division
Rather than work through all 12 possibilities, I will do one that works.
[tex]\begin{array}{r|rrrr}-2 & 2 & 6 & -1 & -10\\& & -4& -4 & 10\\& 2 & 2& -5 & 0\\\end{array}[/tex]
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.
