Problem 64 - Algebra for College Students

Question

Find all real number solutions for each equation. $$4 x^{3}=64 x$$

Step-by-Step Solution

Answer

The solutions are \(x = 0, x = 4, \text{ and } x = -4\).

Step 1: Identify the Equation Type

The given equation is \(4x^3 = 64x\). This is a polynomial equation that can be solved by factoring.

Step 2: Simplify the Equation

Divide both sides by 4, which simplifies the equation to: \(x^3 = 16x\).

Step 3: Rearrange the Equation

Move all the terms to one side of the equation to set it equal to zero: \(x^3 - 16x = 0\).

Step 4: Factor the Equation

Factor out the greatest common factor, which is \(x\): \(x(x^2 - 16) = 0\).

Step 5: Further Factorization

Notice that \(x^2 - 16\) is a difference of squares. Factor it as \((x - 4)(x + 4)\). The equation now is \(x(x-4)(x+4) = 0\).

Step 6: Solve Each Factor Equation

Set each factor equal to zero and solve for \(x\):1. \(x = 0\)2. \(x - 4 = 0\) gives \(x = 4\)3. \(x + 4 = 0\) gives \(x = -4\)

Step 7: Verify the Solutions

Substitute \(x = 0, x = 4, \text{ and } x = -4\) back into the original equation \(4x^3 = 64x\) to ensure they satisfy the equation.- For \(x = 0\): \(4(0)^3 = 64(0)\), both sides are 0.- For \(x = 4\): \(4(4)^3 = 64(4)\), both sides are 256.- For \(x = -4\): \(4(-4)^3 = 64(-4)\), both sides are -256.

Key Concepts

Factoring Polynomials

When solving polynomial equations, factoring is often a crucial first step. In our exercise, we start with the equation \(4x^3 = 64x\). The goal here is to simplify the polynomial equation so that it can be set to zero, making it easier to solve.

• The first step involves simplifying the equation by removing common factors on both sides. Here, dividing by 4 simplifies our expression to \(x^3 = 16x\).
• Next, we rearrange the equation to \(x^3 - 16x = 0\). This is crucial because a polynomial equation set to zero can be solved by finding its roots.
• Factoring involves breaking down the expression into simpler polynomials. We notice that \(x(x^2 - 16) = 0\) can further be factored due to the difference of squares, which we'll explore further in the next section.

The process of factoring transforms the expression into a form where the solutions for the polynomial can be identified as the values that make any of the factors equal to zero.

Difference of Squares

The concept of the difference of squares often appears in polynomial equations. It refers to expressions of the form \(a^2 - b^2\), which can be factored into \((a - b)(a + b)\).In the exercise, once we factor out \(x\) from \(x^3 - 16x\), we are left with \(x^2 - 16\). This is a perfect example of the difference of squares, where \(a = x\) and \(b = 4\).

• Applying the difference of squares formula, \(x^2 - 16\) becomes \((x - 4)(x + 4)\).
• This factorization simplifies the polynomial, allowing us to write the equation as \(x(x - 4)(x + 4) = 0\).

Recognizing and applying the difference of squares rule provides a straightforward pathway to simplify and solve equations that include squared terms.

Solution Verification

After finding the solutions, it's important to verify that they indeed satisfy the original equation. Verification ensures that no mistakes were made during factorization and solving steps.

• We solved the equation \(x(x-4)(x+4) = 0\) and found the potential solutions: \(x = 0\), \(x = 4\), and \(x = -4\).
• Verification involves substituting each potential solution back into the original equation \(4x^3 = 64x\).
• If each substitution results in a true statement, then the solutions are verified:
• For \(x = 0\): \(4(0)^3 = 64(0)\) yields 0 = 0, which is true.
• For \(x = 4\): \(4(4)^3 = 64(4)\) yields 256 = 256, which is true.
• For \(x = -4\): \(4(-4)^3 = 64(-4)\) yields -256 = -256, which is true.

This final step of verification confirms that all our provided solutions are accurate and correct for the given polynomial equation.