Answer:
![\boxed{(1) \, x = \, \pm \dfrac{1}{2}, \pm 1, \pm2, \pm \dfrac{5}{2}, \pm 5, \pm 10; (2) \, x = -2}](https://tex.z-dn.net/?f=%5Cboxed%7B%281%29%20%5C%2C%20x%20%3D%20%5C%2C%20%5Cpm%20%5Cdfrac%7B1%7D%7B2%7D%2C%20%5Cpm%201%2C%20%5Cpm2%2C%20%5Cpm%20%5Cdfrac%7B5%7D%7B2%7D%2C%20%5Cpm%205%2C%20%5Cpm%2010%3B%20%282%29%20%5C%2C%20x%20%3D%20-2%7D)
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.
![\text{Possible rational root} = \dfrac{ p }{ q } = \dfrac{\text{factor of constant term}}{\text{factor of leading coefficient}}](https://tex.z-dn.net/?f=%5Ctext%7BPossible%20rational%20root%7D%20%3D%20%5Cdfrac%7B%20p%20%7D%7B%20q%20%7D%20%3D%20%5Cdfrac%7B%5Ctext%7Bfactor%20of%20constant%20term%7D%7D%7B%5Ctext%7Bfactor%20of%20leading%20coefficient%7D%7D)
In your function, the constant term is -10 and the leading coefficient is 2, so
![\text{Possible root} = \dfrac{\text{factor of 10}}{\text{factor of 2}}](https://tex.z-dn.net/?f=%5Ctext%7BPossible%20root%7D%20%3D%20%5Cdfrac%7B%5Ctext%7Bfactor%20of%2010%7D%7D%7B%5Ctext%7Bfactor%20of%202%7D%7D)
Factors of 10 = ±1, ±2, ±5, ±10
Factors of 2 = ±1, ±2
![\text{Possible roots are } \large \boxed{\mathbf{x = \pm \dfrac{1}{2}, \pm 1, \pm2, \pm \dfrac{5}{2}, \pm 5, \pm 10}}](https://tex.z-dn.net/?f=%5Ctext%7BPossible%20roots%20are%20%7D%20%5Clarge%20%5Cboxed%7B%5Cmathbf%7Bx%20%3D%20%5Cpm%20%5Cdfrac%7B1%7D%7B2%7D%2C%20%5Cpm%201%2C%20%5Cpm2%2C%20%5Cpm%20%5Cdfrac%7B5%7D%7B2%7D%2C%20%5Cpm%205%2C%20%5Cpm%2010%7D%7D)
(2) Synthetic division
Rather than work through all 12 possibilities, I will do one that works.
![\begin{array}{r|rrrr}-2 & 2 & 6 & -1 & -10\\& & -4& -4 & 10\\& 2 & 2& -5 & 0\\\end{array}](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%7Br%7Crrrr%7D-2%20%26%202%20%26%206%20%26%20-1%20%26%20-10%5C%5C%26%20%26%20-4%26%20-4%20%26%2010%5C%5C%26%202%20%26%202%26%20-5%20%26%200%5C%5C%5Cend%7Barray%7D)
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.