Let f(x) = x³ + 4x² + x - 6 - Edexcel - A-Level Maths Pure - Question 7 - 2007 - Paper 2

Given that (x + 2) is a factor, we can divide f(x) by (x + 2). Performing polynomial long division:

1. Divide the leading term: $\frac{x^3}{x} = x^2$.
2. Multiply (x + 2) by x² giving $x³ + 2x^2$.
3. Subtract from f(x): $f(x) - (x³ + 2x²) = 2x² + x - 6$.
4. Repeat the process: Divide $2x²$ by $x$ yielding $2x$.
5. Multiply (x + 2) by $2x$ to get $2x² + 4x$ and subtract: $2x² + x - 6 - (2x² + 4x) = -3x - 6.$
6. Finally, divide $-3x$ by $x$ to get $-3$ and complete the division: $-3(x + 2) = -3x - 6.$
   Thus, we have: $f(x) = (x + 2)(x^2 + 2x - 3).$
   Now, we factor $x^2 + 2x - 3$: $x^2 + 2x - 3 = (x + 3)(x - 1).$
   Therefore, the complete factorization is: $f(x) = (x + 2)(x + 3)(x - 1).$

To find the solutions, we set f(x) = 0: $(x + 2)(x + 3)(x - 1) = 0$. The solutions are found by setting each factor equal to zero:

1. $x + 2 = 0$
   => $x = -2$
2. $x + 3 = 0$
   => $x = -3$
3. $x - 1 = 0$
   => $x = 1$. Thus, the solutions to the equation are: $x = -2, -3, 1.$