Problem 19 - Algebra 2

Question

Solve each equation. $$ 8 x^{3}-1=0 $$

Step-by-Step Solution

Answer

The real solution is \( x = 1/2 \) because the quadratic factor has no real roots.

Step 1: Recognize the Type of Equation

Identify that the equation is a difference of cubes because it can be written as \( a^3 - b^3 = 0 \), where \( a = 2x \) and \( b = 1 \) since \( (2x)^3 = 8x^3 \) and \( 1^3 = 1 \) respectively.

Step 2: Apply the Difference of Cubes Formula

Use the difference of cubes formula \( a^3 - b^3 = (a - b)(a^2 + ab + b^2) \) to factor the equation. Replace \( a \) and \( b \) with \( 2x \) and \( 1 \) to get \( (2x - 1)[(2x)^2 + (2x)(1) + 1^2] = 0 \) which simplifies to \( (2x - 1)(4x^2 + 2x + 1) = 0 \).

Step 3: Find the Real Solutions

Set each factor equal to zero and solve for \( x \). From \( 2x - 1 = 0 \) we get \( x = 1/2 \). The quadratic equation \( 4x^2 + 2x + 1 = 0 \) does not have real solutions because the discriminant \( b^2 - 4ac = 2^2 - 4(4)(1) = 4 - 16 = -12 \) is negative. Therefore, there is only one real solution: \( x = 1/2 \) which comes from the linear factor \( 2x - 1 = 0 \).

Key Concepts

Difference of Cubes Formula

The difference of cubes refers to the subtraction of two cubed numbers, symbolized as \( a^3 - b^3 \). This expression can be factored into a binomial times a trinomial using a specific formula: \( a^3 - b^3 = (a - b)(a^2 + ab + b^2) \).

For anyone grappling with polynomial equations, understanding this formula is a game-changer. It helps to transform a cubic polynomial, which can be quite daunting to solve, into a more manageable product of factors. Let's dive into our example: \( 8x^3 - 1 = 0 \). By recognizing that \( 8x^3 \) is the cube of \( 2x \) and that \( 1 \) is the cube of \( 1 \), we can apply our formula, replacing \( a \) with \( 2x \) and \( b \) with \( 1 \) to get \( (2x - 1)(4x^2 + 2x + 1) = 0 \).

By factoring out the given polynomial using this formula, we set the stage for finding the equation's real solutions with greater ease.

Factoring Polynomials

Factoring polynomials is akin to breaking down numbers into their prime components; it's about finding the simplest expressions that multiply together to form the original polynomial. It can drastically simplify the process of solving algebraic equations.

Factoring may involve pulling out a greatest common factor, grouping terms, or utilizing formulas like the difference of cubes as we've previously seen. In our current scenario, after applying the difference of cubes formula, we obtained two factors: a linear factor \( (2x - 1) \) and a quadratic factor \( (4x^2 + 2x + 1) \).

To solve for \( x \), we set each factor equal to zero. For the linear factor \( 2x - 1 = 0 \), solving gives us the real solution \( x = 1/2 \). However, for the quadratic factor, no such real solutions exist, which we determine using the concept of discriminants.

Real Solutions of Equations

When we talk about finding real solutions to equations, we're searching for the x-values that make the equation true when we substitute them back in. They are the points where the function crosses the x-axis on a graph.

In polynomial equations, these solutions often come directly from the factors set to zero. However, quadratic factors present a unique challenge. The discriminant—given by \( b^2 - 4ac \) in the quadratic formula—helps us predict the number of real solutions. If it's positive, we have two real solutions; if zero, one real solution; and if negative, no real solutions. Back to our equation, the quadratic factor produced a negative discriminant, and therefore, it confirmed no real solutions for that portion of the equation.

In summary, our original cubic equation had only one real solution from its linear factor. Understanding how to factor polynomials, and knowing the behavior of discriminants, is essential in successfully solving equations and finding their real solutions.