Problem 51 - Basic Technical Mathematics with Calculus

Question

Although the equations are not quadratic, factoring will lead to one quadratic factor and the solution can be completed by factoring as with a quadratic equation. Find the three roots of each equation. $$x^{3}-x=0$$

Step-by-Step Solution

Answer

The roots of the equation are \( x = 0, x = 1, \) and \( x = -1 \).

Step 1: Identify the Common Factor

Given the equation \( x^3 - x = 0 \), notice that we have a common factor of \( x \) on both terms. This means we can factor \( x \) out of the equation. Write the equation in its factored form as follows: \( x(x^2 - 1) = 0 \).

Step 2: Factor the Quadratic Expression

The quadratic expression \( x^2 - 1 \) is a difference of two squares, which can be factored into \( (x - 1)(x + 1) \). Therefore, the entire expression becomes \( x(x - 1)(x + 1) = 0 \).

Step 3: Solve for Each Factor Equal to Zero

Set each factor equal to zero to find the solutions for \( x \):1. \( x = 0 \)2. \( x - 1 = 0 \Rightarrow x = 1 \)3. \( x + 1 = 0 \Rightarrow x = -1 \)

Step 4: List All the Roots

The solutions to the equation \( x^3 - x = 0 \) are the values of \( x \) that make any of the factors zero. Thus, the roots of the equation are \( x = 0, x = 1, \) and \( x = -1 \).

Key Concepts

Common Factor

In algebra, identifying a common factor is key in simplifying and solving equations effectively. A common factor is a number or variable that divides each term in an expression evenly. For example, in the equation \( x^3 - x = 0 \), both terms, \( x^3 \) and \( x \), share \( x \) as a common factor.
Factoring out the common term simplifies the equation, making it easier to solve. In this case, factoring \( x \) from the equation gives us \( x(x^2 - 1) = 0 \). This new expression is much simpler and paves the way for further factorization. Finding common factors helps reduce polynomials, setting the stage for techniques like the difference of squares.

Difference of Squares

The difference of squares is another factorization method widely used in algebra. This technique applies to expressions of the form \( a^2 - b^2 \), which can be rewritten as \( (a - b)(a + b) \). It hinges on the fact that the squares subtracting each other have a straightforward pattern.
In our given equation, after factoring out the \( x \), we have \( x^2 - 1 \) remaining. This expression is actually a difference of squares, because it equals \( x^2 - 1^2 \). We can factor it further to become \( (x - 1)(x + 1) \).
Recognizing the difference of squares allows us to break expressions down into simpler linear factors, making it easier to find solutions.

Polynomial Roots

Finding the roots of a polynomial means identifying all values for which the polynomial equals zero. These are the solutions to the equation.
With the expression \( x(x - 1)(x + 1) = 0 \) fully factored, we can find the roots by setting each factor equal to zero independently. This gives us three separate equations:

• \( x = 0 \)
• \( x - 1 = 0 \), which solves to \( x = 1 \)
• \( x + 1 = 0 \), which solves to \( x = -1 \)

These values, \( x = 0, 1, -1 \), are the roots of the original equation \( x^3 - x = 0 \). Identifying roots is fundamental in understanding the behavior of polynomials and their graphical representations.