Problem 1 - A Graphical Approach to Precalculus with Limits

Question

Find all real solutions. Do not use a calculator. $$x^{3}-25 x=0$$

Step-by-Step Solution

Answer

The solutions are \(x = 0\), \(x = 5\), and \(x = -5\).

Step 1: Factor the Equation

The given equation is \(x^3 - 25x = 0\). Start by factoring out the greatest common factor, which is \(x\). This gives: \(x(x^2 - 25) = 0\).

Step 2: Solve for x from the Factored Terms

Now solve each factor separately: 1. \(x = 0\) is one solution.2. \(x^2 - 25 = 0\) can be solved by recognizing it as a difference of squares.Let's solve \(x^2 - 25 = 0\).

Step 3: Factor the Difference of Squares

The equation \(x^2 - 25 = 0\) can be factored as \((x - 5)(x + 5) = 0\). Now solve each binomial factor for \(x\).

Step 4: Solve the Binomial Equations

Set each binomial equal to zero and solve for \(x\):1. \(x - 5 = 0\) gives \(x = 5\).2. \(x + 5 = 0\) gives \(x = -5\).

Step 5: List All Solutions

Having solved all the factors, the real solutions are obtained. Thus, the solutions are \(x = 0\), \(x = 5\), and \(x = -5\).

Key Concepts

Solving Polynomial Equations

When we talk about solving polynomial equations, we mean finding values for the variable that make the equation true. A polynomial equation is composed of terms which have variables raised to whole number powers, like in our example, \(x^3 - 25x = 0\). To solve, we want to find "roots" or "solutions". These solutions are the values of \(x\) making the polynomial equal to zero.

Solving these requires:

• Identifying the polynomial type (linear, quadratic, cubic, etc.).
• Deciding a method to solve it, like factoring, using the quadratic formula, or graphing.

In our case, factoring is a good start, as it breaks down the polynomial into simpler expressions. Once factored, each smaller expression can be set to zero to find possible solutions. This is an efficient and powerful way to tackle polynomial equations, especially when higher-degree polynomials are involved.

Factoring

Factoring is like breaking down a big puzzle into smaller, more manageable pieces. To factor a polynomial means to express it as a product of its factors. For our polynomial \(x^3 - 25x = 0\), we noticed the greatest common factor is \(x\). By factoring out \(x\), we simplify the equation to \(x(x^2 - 25) = 0\).

There are several techniques for factoring:

• Greatest Common Factor (GCF): Like we did, find the largest factor common to all terms.
• Factoring by Grouping: Group terms that can factor out a common term.
• Special Formulas: Use special factoring formulas for quadratics or recognizing sums or differences of cubes or squares.

Factoring simplifies solving. Each factor is set to zero, transforming complex polynomials into solvable equations. With practice, recognizing what to factor becomes second nature, enabling quick and efficient solutions.

Difference of Squares

The difference of squares is a powerful factoring technique. It applies to expressions of the form \(a^2 - b^2\). It's known as a "difference of squares" because it involves two squares being subtracted. Our expression \(x^2 - 25 = 0\) is a textbook example where \(x^2\) and 25 are perfect squares.

The formula for factoring a difference of squares is:

• \(a^2 - b^2 = (a - b)(a + b)\)

Applying this to \(x^2 - 25\), we set \(x^2\) as \(a^2\) and 25 as \(b^2\), yielding \((x - 5)(x + 5) = 0\). This allows us to solve \(x - 5 = 0\) giving \(x = 5\), and \(x + 5 = 0\) giving \(x = -5\).

Recognizing and utilizing the difference of squares swiftly simplifies solving polynomial equations and provides the necessary solutions efficiently.