Problem 47 - Precalculus: Functions and Graphs

Question

Find the solutions of the equation. $$x^{3}+125=0$$

Step-by-Step Solution

Answer

The real solution is \(x = -5\).

Step 1: Identify the Structure of the Equation

The given equation is in the form \(x^3 + 125 = 0\). This is a cubic equation with no terms in \(x^2\) or \(x\).

Step 2: Rearrange the Equation

Subtract 125 from both sides of the equation to isolate the cubic term: \(x^3 = -125\).

Step 3: Identify as a Cubic Difference of Two Cubes

Recognize the equation as a difference of cubes: \(x^3 = (-5)^3\), since \(-125\) is the cube of \(-5\).

Step 4: Factor the Difference of Cubes

Use the formula for factoring the difference of cubes, \(a^3 - b^3 = (a-b)(a^2+ab+b^2)\). Here, \(a = x\) and \(b = -5\), so the equation becomes: \((x + 5)(x^2 - 5x + 25) = 0\).

Step 5: Solve for Real Solutions

Set each factor in the equation to zero: - \(x + 5 = 0\) gives \(x = -5\). - The quadratic \(x^2 - 5x + 25\) has no real solutions since the discriminant \((-5)^2 - 4 \times 1 \times 25 = -75\) is negative.

Key Concepts

Factoring Polynomials

Factoring polynomials is like breaking down a big problem into smaller, more manageable pieces. It's similar to breaking down a number into its prime factors. With polynomials, factorization helps us to simplify equations and find the solutions more easily. For example, if you have a polynomial like \(x^3 + 125\), you can factor it by recognizing it as a sum of cubes. This involves expressing the polynomial in terms of its factors, such as

• \((x + 5)\)
• and \((x^2 - 5x + 25)\).

Once factored, solving becomes straightforward because you can tackle each factor separately, setting them to zero to find possible solutions. This is a fundamental skill in algebra that makes solving more complex equations manageable and easy to understand.

Difference of Cubes

The difference of cubes is a special technique for factoring certain types of cubic equations. In equations like \(x^3 - y^3\), the difference of cubes can be applied. This involves rewriting the expression using the formula:

• \(a^3 - b^3 = (a-b)(a^2 + ab + b^2)\).

In our exercise, recognizing that \(x^3 + 125\) can be rewritten as \(x^3 - (-5)^3\), allows us to apply this method. Applying the difference of cubes formula here transforms the original cubic equation into \((x + 5)(x^2 - 5x + 25)\). By utilizing this formula, complex polynomial expressions are simplified, making it easier to find solutions to the equation by examining each factor individually.

Quadratic Equation Solutions

Solving quadratic equations often involves finding the roots or solutions of the equation. When you break a polynomial equation like \((x^2 - 5x + 25)\), it's essentially a quadratic equation that you need to solve. Quadratic equations can sometimes be solved easily by factoring. However, when factoring is not possible, other methods such as using the quadratic formula are employed.
To determine if a quadratic equation has real solutions, we check its discriminant, \(b^2 - 4ac\). For the equation \(x^2 - 5x + 25\), the discriminant is \(-75\). A negative discriminant implies that the quadratic has no real solutions. Instead, it would be solved over the complex numbers. Recognizing the nature of the solutions through the discriminant is crucial as it tells whether we have real or complex roots.