Problem 52 - Basic Technical Mathematics with Calculus

Question

Although the equations are not quadratic, factoring will lead to one quadratic factor and the solution can be completed by factoring as with a quadratic equation. Find the three roots of each equation. $$x^{3}-4 x^{2}-x+4=0$$

Step-by-Step Solution

Answer

The roots of the equation are \(x = 4\), \(x = 1\), and \(x = -1\).

Step 1: Group Terms

First, group terms in the polynomial to make it easier to factor. Group the first two terms together and the last two terms together:\( (x^3 - 4x^2) + (-x + 4) = 0 \)

Step 2: Factor by Common Factors

Factor out the greatest common factor from each group. This will simplify the expression:\( x^2(x - 4) - 1(x - 4) = 0 \)

Step 3: Apply the Distributive Law

Notice that \((x - 4)\) is a common factor in the two groups, so factor \((x - 4)\) out of the entire expression:\( (x - 4)(x^2 - 1) = 0 \)

Step 4: Factor the Quadratic

The expression \(x^2 - 1\) is a difference of squares, which can be factored further:\( x^2 - 1 = (x - 1)(x + 1) \)

Step 5: Combine Factors for the Full Equation

Substitute the factored quadratic back into the equation:\( (x - 4)(x - 1)(x + 1) = 0 \)

Step 6: Find the Roots

Solve each factor equal to zero to find the roots of the equation:1. \( x - 4 = 0 \) gives \( x = 4 \) 2. \( x - 1 = 0 \) gives \( x = 1 \) 3. \( x + 1 = 0 \) gives \( x = -1 \)

Key Concepts

Factoring

Factoring involves breaking down an expression into a product of simpler expressions or factors. In the original exercise, we start by dividing the polynomial into groups. This method is termed "grouping," where terms are combined to make factoring easier. Here, the equation

• \( x^3 - 4x^2 - x + 4 = 0 \)

is first grouped as

• \( (x^3 - 4x^2) + (-x + 4) \).

By factoring out the greatest common factor from each group, you simplify the expression to reveal hidden common factors. For instance, in the grouped expression:

• First group: \( x^2(x - 4) \)
• Second group: \(-1(x - 4) \)

Here, \( x - 4 \) appears in both sections, allowing it to be factored out using the distributive property. Patterns like these make factoring a foundational technique in solving polynomial equations.

Polynomial Equations

Polynomial equations are expressions involving variables raised to various powers. They appear often in mathematics, reflecting a wide array of real-world phenomena. In simple terms, a polynomial is composed of terms, where each term is a product of a constant and a variable raised to a non-negative integer power. The given problem features a cubic polynomial, meaning its highest degree is three.To solve polynomial equations, factoring them into simpler components often helps, as seen in the exercise.For example, the original equation is

• \( x^3 - 4x^2 - x + 4 = 0 \),

which can be factored into:

• \( (x - 4)(x - 1)(x + 1) = 0 \).

Each of these factors can be solved individually to find the roots, which are the solutions to the equation. This illustrates how understanding and manipulating polynomials through factoring can directly lead to solutions.

Difference of Squares

The difference of squares is a special factoring pattern that occurs when two square terms are subtracted. It has the general form

• \( a^2 - b^2 = (a - b)(a + b) \).

In the context of the provided solution, the term

• \( x^2 - 1 \)

is recognized as a difference of squares since

• \( 1 \) can be rewritten as \( (1)^2 \).

Thus, it can be factored accordingly into

• \( (x - 1)(x + 1) \).

This pattern enables simpler factoring of quadratic expressions, providing a quick and reliable factorization method. Mastering the difference of squares concept gives you an edge in efficiently solving polynomial equations, especially when encountering quadratic terms that fit this recognizable pattern.