Solve the following equation in the real number system. Please show all of your work.

2x^3-3x^2-14x+15=0

I can see right off that x = 1 works

2-3-14+15 = 0 sure enough
so divide by (x-1)--> 2x^2-x-1
(2x+1)(x-1)
so x = 1 again and x = -1/2

To solve the equation 2x^3 - 3x^2 - 14x + 15 = 0, we will use the Rational Root Theorem and synthetic division to find the roots.

First, we need to list all the possible rational roots of the equation. The rational roots will be in the form p/q, where p is a factor of 15 (the constant term) and q is a factor of 2 (the leading coefficient).

The factors of 15 are ±1, ±3, ±5, and ±15.
And the factors of 2 are ±1 and ±2.

Hence, the possible rational roots are: ±1/1, ±3/1, ±5/1, ±15/1, ±1/2, ±3/2, ±5/2, ±15/2.

Now we will perform synthetic division to test these possible roots one by one.

Testing x = 1:
1 | 2 -3 -14 15
| 2 -1 -15
¯¯¯¯¯¯¯¯¯¯¯¯
2 -1 -15 0
Since the remainder is zero, x = 1 is a root.

Now, we need to solve the equation (2x^3 - 3x^2 - 14x + 15 = 0)/(x - 1) = 0 by factoring. We can use long division or rewrite the equation as a product of polynomials.

Dividing 2x^3 - 3x^2 - 14x + 15 by (x - 1), we get:
(2x^3 - 3x^2 - 14x + 15)/(x - 1) = 2x^2 - x - 15

Now we can factor this quadratic equation:
2x^2 - x - 15 = 0
(2x + 5)(x - 3) = 0

Setting each factor equal to zero:
2x + 5 = 0 or x - 3 = 0

Solving for x:
2x = -5 or x = 3

Dividing 2x + 5 = 0 by 2, we get:
x = -5/2

Therefore, the roots of the equation 2x^3 - 3x^2 - 14x + 15 = 0 in the real number system are:
x = 1, x = -5/2, x = 3.

Hence, the solution to the equation is x = 1, -5/2, and 3.