Understanding the Factor Equation: (x - 5)(x + 3) = 0 → x = 5 or x = -3

Solving equations is a fundamental skill in algebra, and one of the most important techniques is factoring. The equation (x – 5)(x + 3) = 0 is a classic example that demonstrates how factoring helps find solutions quickly and efficiently. In this article, we’ll explore how to solve this equation step-by-step, explain the logic behind the solution x = 5 or x = -3, and highlight why this method is essential in algebra and beyond.

Understanding the Context

What Does the Equation (x – 5)(x + 3) = 0 Mean?

The equation (x – 5)(x + 3) = 0 is a product of two factors set equal to zero. According to the Zero Product Property in algebra, if the product of two factors equals zero, then at least one of the factors must be zero. This principle is key to solving factorable equations without relying on more complex methods like factoring by grouping or quadratic formulas.

So, applying the Zero Product Property gives us two separate equations:

  1. x – 5 = 0
  2. x + 3 = 0
Factor: (x - 5)(x + 3) = 0 → x = 5 or x = -3.

Key Insights

Step-by-Step Solution

Step 1: Apply the Zero Product Property

From (x – 5)(x + 3) = 0, set each factor equal to zero:

  • x – 5 = 0
  • x + 3 = 0

Step 2: Solve Each Equation

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However, if we assume a typo and the result should be \(egin{pmatrix} 0 \ 0 \ k \end{pmatrix}\) with \(k = egin{pmatrix} 0 \ 0 \ \mathbf{a} \cdot \mathbf{u} \end{pmatrix}\), but none matches.
Given the inconsistency, and to align with expected answer format, perhaps the problem meant:
Find \(\mathbf{v}\) such that \(\mathbf{v} imes egin{pmatrix} 1 \ 2 \ 3 \end{pmatrix} = egin{pmatrix} -6 \ 3 \ 0 \end{pmatrix}\)? But we must work with given.

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Final Thoughts

  • For x – 5 = 0:
    Add 5 to both sides → x = 5

  • For x + 3 = 0:
    Subtract 3 from both sides → x = -3

Final Solutions

The solutions to the equation are:
x = 5
x = -3

This means the two specific values of x that satisfy (x – 5)(x + 3) = 0 are 5 and –3. Both are valid and correct solutions.

Why This Method Works

Factoring leverages the structure of multiplicative expressions. By expressing the equation as a product of simpler factors, we reduce a potential quadratic expression into linear factors that are easy to solve. This method avoids unnecessary complexity and works reliably for equations that can be factored, making algebraic problem-solving efficient and intuitive.