Problem 33 - Algebra 2

Question

Solve each equation. $$ x^{3}+64=0 $$

Step-by-Step Solution

Answer

The real solution is \( x = -4 \).

Step 1: Recognize the Equation Type

The equation given is a polynomial equation of degree 3: \( x^3 + 64 = 0 \). Notice that 64 is a perfect cube (since \( 4^3 = 64 \)), which suggests that the equation could be solved by factoring as a sum of cubes.

Step 2: Use the Sum of Cubes Formula

The sum of cubes formula is: \( a^3 + b^3 = (a + b)(a^2 - ab + b^2) \). Here, \( a = x \) and \( b = 4 \), because \( 4^3 = 64 \). Therefore, the equation can be rewritten as \( (x + 4)(x^2 - 4x + 16) = 0 \).

Step 3: Solve Each Factor

Set each factor equal to zero. For the first factor, \( x + 4 = 0 \). Solving gives \( x = -4 \).The second factor, \( x^2 - 4x + 16 = 0 \), is a quadratic equation. To find its roots, calculate the discriminant: \( b^2 - 4ac = (-4)^2 - 4 \cdot 1 \cdot 16 = 16 - 64 = -48 \). A negative discriminant means there are no real roots for this factor.

Step 4: Determine the Real Solution

From Step 3, the only real solution comes from the factor \( x + 4 = 0 \), which gives the solution \( x = -4 \). The quadratic factor \( x^2 - 4x + 16 = 0 \) has no real solutions due to the negative discriminant.

Key Concepts

Sum of Cubes

The sum of cubes formula is a powerful tool for factoring polynomial equations involving cube terms. The formula is expressed as \(a^3 + b^3 = (a + b)(a^2 - ab + b^2)\). This formula is essential because it helps break down complex cubic expressions into simpler parts that are easier to solve or analyze.

When using the sum of cubes formula, it's crucial to identify the components correctly, which in this case are the terms \(a^^{3}\) and \(b^^{3}\). Following our exercise, we have \(x^3 + 64\), where 64 is recognized as \(b^3\), because \(4^3 = 64\).

With these identified, the application of the sum of cubes formula allows us to express the polynomial as a product of two factors: \(x + 4\) and \(x^2 - 4x + 16\). This transformation makes it possible to progress towards solving the equation.

Factoring Polynomials

Factoring polynomials, especially cubics, is a method that simplifies them into products of polynomials of lower degrees, ideally leading to easier solutions. In the provided exercise, factoring was achieved using the sum of cubes formula, which split \(x^3 + 64\) into \(x + 4\) and \(x^2 - 4x + 16\).

Factoring is a vital technique because it often reveals simple roots, as seen with the factor \(x + 4 = 0\), which directly gives us a real solution \(x = -4\).

The second factor, \(x^2 - 4x + 16\), being a quadratic equation, requires additional steps like checking the discriminant to find roots, which in this case, did not yield real numbers due to a negative discriminant.

Complex Roots

When solving polynomial equations, encountering complex roots is common, especially in situations where a quadratic equation has a negative discriminant. Complex roots arise from such quadratics through the formula \((-b \pm \sqrt{b^2 - 4ac})/(2a)\), where the term under the square root, called the discriminant, dictates the root nature.

In the exercise, the equation \(x^2 - 4x + 16 = 0\) possesses a discriminant of \(-48\), indicating the roots are complex. Complex roots take the form \(p \pm qi\), where \(i\) represents the imaginary unit\((i^2 = -1)\).

Understanding complex roots is essential because polynomial equations can have no real roots but still be fully solvable within the complex number system, ensuring a complete solution set.