Problem 77 - Algebra for College Students

Question

Should help you pull together all of the factoring techniques of this chapter. Factor completely each polynomial, and indicate any that are not factorable using integers. $$n^{3}-49 n$$

Step-by-Step Solution

Answer

The polynomial \(n^3 - 49n\) factors to \(n(n - 7)(n + 7)\).

Step 1: Identify the Greatest Common Factor (GCF)

First, determine if there is a greatest common factor in the polynomial. The terms in the polynomial are \(n^3\) and \(-49n\), both of which have \(n\) in common. Therefore, the GCF is \(n\).

Step 2: Factor Out the GCF

Factor the greatest common factor (\(n\)) out of each term in the polynomial: \[n(n^2 - 49)\].

Step 3: Recognize a Difference of Squares

Observe that \(n^2 - 49\) is a difference of squares. Recall that a difference of squares can be expressed as \(a^2 - b^2 = (a - b)(a + b)\). Here, \(a = n\) and \(b = 7\) since \(49 = 7^2\).

Step 4: Apply the Difference of Squares Formula

Use the difference of squares formula to factor \(n^2 - 49\): \[n^2 - 49 = (n - 7)(n + 7)\].

Step 5: Combine the Factored Parts

Combine the factored parts from the previous steps to express the original polynomial entirely factored:\[n(n - 7)(n + 7)\].

Key Concepts

Greatest Common Factor

In algebra, finding the greatest common factor (GCF) of terms within a polynomial is often the first step in the factoring process. The GCF is the largest expression that can be evenly divided into each term of the polynomial. It can significantly simplify further factoring processes.

To find the GCF of a polynomial like \(n^3 - 49n\), follow these steps:

• Identify common factors in each term. Both \(n^3\) and \(-49n\) contain the variable \(n\).
• Determine the highest power of the common factor present in each term. The smallest power of \(n\) across the terms is \(n^1\).

Thus, the GCF of \(n^3 - 49n\) is \(n\). Factoring this out simplifies the polynomial to \(n(n^2 - 49)\). This step makes other factoring techniques easier to apply later on.

Difference of Squares

The difference of squares is a special factoring technique used when a polynomial is structured as \(a^2 - b^2\). This unique form allows straightforward factoring into two binomials: \((a - b)(a + b)\).

After factoring out the GCF, the expression \(n^2 - 49\) from the polynomial \(n(n^2 - 49)\) fits this pattern because:

• \(n^2\) is the square of \(n\); hence \(a = n\).
• \(49\) is the square of \(7\); thus \(b = 7\).

The expression \(n^2 - 49\) is essentially \((n^2 - 7^2)\), a classic difference of squares. Applying the formula, it factors into \((n - 7)(n + 7)\). This technique is a powerful tool as it simplifies expressions that seem complex at first glance.

Factoring Techniques

Factoring is a crucial method in algebra used to break down polynomials into simpler parts, thus making them easier to solve or simplify. Several techniques exist, each applicable under specific conditions:

• Greatest Common Factor (GCF): Always the first step, identify and factor out the GCF as seen with \(n(n^2 - 49)\).
• Difference of Squares: If a polynomial is structured as \(a^2 - b^2\), use \((a - b)(a + b)\), which was applied to \(n^2 - 49\) to obtain \((n - 7)(n + 7)\).
• Trinomials: Common polynomials of the form \(ax^2 + bx + c\), where strategies like splitting the middle term or the quadratic formula are used.
• Grouping: Useful when polynomials have more than three terms, often involving rearranging and factoring out common factors in groups.

By combining these techniques, like using the GCF and difference of squares as shown, polynomials can be systematically reduced to a product of simpler terms. This process not only aids in solving equations but also in understanding the structure and solutions of algebraic expressions.