Given P(x) = 2x³ + 5x² - x - 6. Using your solution to P(x) = 0, write down the solution of the equation P( x + 1) = 0

Given the polynomial \( P(x) = 2x^3 + 5x^2 - x - 6 \), we want to first solve the equation \( P(x) = 0 \) and then use that solution to find the solution to \( P(x + 1) = 0 \).

### Step 1: Solve \( P(x) = 0 \)
We need to solve the cubic equation:
\[
2x^3 + 5x^2 - x - 6 = 0
\]

This is a cubic equation, and solving it generally involves checking for rational roots using the Rational Root Theorem or using numerical methods. Let's apply the Rational Root Theorem.

Possible rational roots are the factors of \(-6\) (constant term) divided by the factors of \(2\) (leading coefficient). The possible rational roots are:
\[
\pm 1, \pm 2, \pm 3, \pm 6, \pm \frac{1}{2}, \pm \frac{3}{2}
\]

Testing these, we find that \( x = 1 \) is a root:
\[
P(1) = 2(1)^3 + 5(1)^2 - 1 - 6 = 2 + 5 - 1 - 6 = 0
\]

So, \( x = 1 \) is a root. We can now factor \( P(x) \) as:
\[
P(x) = (x - 1)(2x^2 + 7x + 6)
\]

The quadratic factor \( 2x^2 + 7x + 6 \) can be factored further:
\[
2x^2 + 7x + 6 = (2x + 3)(x + 2)
\]

Thus, the full factorization of \( P(x) \) is:
\[
P(x) = (x - 1)(2x + 3)(x + 2)
\]

The solutions to \( P(x) = 0 \) are:
\[
x = 1, \quad x = -\frac{3}{2}, \quad x = -2
\]

### Step 2: Solve \( P(x + 1) = 0 \)
We replace \( x \) with \( x + 1 \) in the original equation:
\[
P(x + 1) = 2(x + 1)^3 + 5(x + 1)^2 - (x + 1) - 6 = 0
\]

Using the roots \( x = 1 \), \( x = -\frac{3}{2} \), and \( x = -2 \) from \( P(x) = 0 \), the solutions to \( P(x + 1) = 0 \) are obtained by solving:
\[
x + 1 = 1 \quad \Rightarrow \quad x = 0
\]
\[
x + 1 = -\frac{3}{2} \quad \Rightarrow \quad x = -\frac{5}{2}
\]
\[
x + 1 = -2 \quad \Rightarrow \quad x = -3
\]

### Final Solution
The solutions to the equation \( P(x + 1) = 0 \) are:
\[
x = 0, \quad x = -\frac{5}{2}, \quad x = -3
\]

Would you like more details or have any questions about this solution?

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Here are five related questions you might explore:
1. How would you factor the cubic polynomial \( P(x) \) if given different coefficients?
2. What is the geometric interpretation of the roots of a cubic polynomial?
3. How does the Rational Root Theorem help in finding solutions to polynomial equations?
4. Can you derive the roots of a quadratic equation using the quadratic formula?
5. How does shifting the variable \( x \) by a constant (as in \( P(x + 1) \)) affect the roots of a polynomial?

**Tip:** When solving cubic equations, always check for possible rational roots first. It simplifies the process significantly if a root is found early.

Learn how to solve the polynomial equation P(x) = 2x^3 + 5x^2 - x - 6 and its transformation P(x + 1) = 0. Detailed step-by-step solution using the Rational Root Theorem and factoring techniques. Suitable for high school students.

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