Solving the Equation y x^3 - 3x^2 - x - 2 Using Algebraic Techniques

The equation in question is:

1. Setting up the Equation

To solve the equation y x^3 - 3x^2 - x - 2 for x when y 0, we begin by setting the equation to zero:

0 x^3 - 3x^2 - x - 2

2. Checking for Rational Roots Using the Rational Root Theorem

The Rational Root Theorem suggests that any rational root of the polynomial is a factor of the constant term (here, -2) divided by the factors of the leading coefficient (here, 1).

Factors of -2: ±1, ±2

Therefore, the possible rational roots are ±1, ±2.

3. Testing Possible Roots

We will test these roots one by one using the equation:

For (x 1):

13 - 3(1)2 - 1 - 2 1 - 3 - 1 - 2 -5 ≠ 0 (not a root)

For (x -1):

(-1)3 - 3(-1)2 - (-1) - 2 -1 - 3 1 - 2 -5 ≠ 0 (not a root)

For (x 2):

23 - 3(2)2 - 2 - 2 8 - 12 - 2 - 2 -8 ≠ 0 (not a root)

For (x -2):

(-2)3 - 3(-2)2 - (-2) - 2 -8 - 12 2 - 2 -20 2 - 2 0 (root)

4. Factoring the Polynomial

Since (x -2) is a root, we can factor the polynomial using synthetic division or long division. Dividing (x^3 - 3x^2 - x - 2) by (x - 2), we get:

Quotient: (x^2 - x - 1)

So, we can rewrite the polynomial as:

x^3 - 3x^2 - x - 2 (x 2)(x^2 - x - 1)

5. Solving the Quadratic Equation

Now, we solve the quadratic equation (x^2 - x - 1 0) using the quadratic formula:

x (-b pm sqrt{b^2 - 4ac}) / (2a)

Here, (a 1), (b -1), and (c -1). Plugging in the values, we get:

x (frac{1 pm sqrt{(-1)^2 - 4(1)(-1)}}{2(1)} frac{1 pm sqrt{1 4}}{2} frac{1 pm sqrt{5}}{2})

6. Summary of Solutions

The solutions to the original equation are:

(x -2) (x frac{1 sqrt{5}}{2} approx 1.618) (x frac{1 - sqrt{5}}{2} approx -0.618)

7. Additional Insights

When simplifying expressions, notice that adding and subtracting terms can sometimes help. For the equation, this manipulation can be useful:

y x^3 - 3x^2 - x - 2 x^3 - 3x^2 - 3x 2x - 1 - 3x - 1

Which simplifies to:

y (x - 1)^3 - 2x - 1

From this form, it's easier to see that when y -1, x -1.

Double-checking results using simpler methods can be a useful way to verify more advanced calculations. If we want (y 0), the equation simplifies to x(1)^3 2x - 1, suggesting that (x -1) when (y 0). This is a simple way to verify the more complex solution.