Find all the roots of the equation. Show your step-by-step work. x^3−x^2−x−2=0
To find all the roots of the equation x^3 - x^2 - x - 2 = 0, we can use the Rational Root Theorem.
The Rational Root Theorem states that if a polynomial equation has a rational root p/q, where p is a factor of the constant term (in this case, -2) and q is a factor of the leading coefficient (in this case, 1), then p/q is a possible root of the equation.
The factors of -2 are ±1 and ±2, and the factors of 1 are ±1. Therefore, the possible rational roots of the equation are ±1 and ±2.
Now, let's try these values one at a time to see if they are roots of the equation.
For x = -1:
(-1)^3 - (-1)^2 - (-1) - 2 = -1 - 1 + 1 - 2 = -3
Since the result is not 0, x = -1 is not a root of the equation.
For x = 1:
(1)^3 - (1)^2 - (1) - 2 = 1 - 1 - 1 - 2 = -3
Since the result is not 0, x = 1 is not a root of the equation.
For x = -2:
(-2)^3 - (-2)^2 - (-2) - 2 = -8 - 4 + 2 - 2 = -10
Since the result is not 0, x = -2 is not a root of the equation.
For x = 2:
(2)^3 - (2)^2 - (2) - 2 = 8 - 4 - 2 - 2 = 0
Since the result is 0, x = 2 is a root of the equation.
Now that we have found one root, x = 2, we can factor the equation using synthetic division or long division. By doing so, we get:
(x - 2)(x^2 + x + 1) = 0
Now let's solve the remaining quadratic equation, x^2 + x + 1 = 0. We can use the quadratic formula:
x = [-b ± √(b^2 - 4ac)] / (2a)
where a = 1, b = 1, and c = 1.
x = [-(1) ± √((1)^2 - 4(1)(1))] / (2(1))
x = [-1 ± √(1 - 4)] / 2
x = [-1 ± √(-3)] / 2
x = [-1 ± √3i] / 2
Therefore, the roots of the equation x^3 - x^2 - x - 2 = 0 are x = 2 and x = (-1 ± √3i)/2.