(x-3)(x^2+2x+1)=5(x+1)

3 min read Jun 17, 2024

Solving the Equation (x-3)(x^2+2x+1) = 5(x+1)

This article will guide you through the process of solving the equation (x-3)(x^2+2x+1) = 5(x+1). We will use algebraic manipulation to isolate x and find its solutions.

Step 1: Expand both sides of the equation

First, we need to expand both sides of the equation. The left side involves multiplying the factors, while the right side requires distributing the 5.

Left Side:
- (x-3)(x^2+2x+1) = x(x^2+2x+1) - 3(x^2+2x+1)
- = x^3 + 2x^2 + x - 3x^2 - 6x - 3
- = x^3 - x^2 - 5x - 3
Right Side:
- 5(x+1) = 5x + 5

This gives us the expanded equation: x^3 - x^2 - 5x - 3 = 5x + 5

Step 2: Move all terms to one side

To solve the equation, we need to set it equal to zero. We achieve this by moving all terms from the right side to the left:

x^3 - x^2 - 10x - 8 = 0

Step 3: Factor the equation

Now we need to factor the cubic equation. We can use different methods for factorization, such as grouping or the Rational Root Theorem. In this case, we can factor by grouping:

Grouping:
- x^3 - x^2 - 10x - 8 = x^2(x-1) - 8(x+1)
- = (x^2-8)(x-1) = 0

Step 4: Solve for x

We have factored the equation into two factors. For the entire expression to be zero, at least one of the factors must be zero. Therefore, we have two possible solutions:

x^2 - 8 = 0
- x^2 = 8
- x = ±√8 = ±2√2
x - 1 = 0
- x = 1

Solution

The solutions to the equation (x-3)(x^2+2x+1) = 5(x+1) are:

x = 1
x = 2√2
x = -2√2