Problem 37 - Elementary and Intermediate Algebra

Question

The following is a list of random factoring problems. Factor each expression. If an expression is not factorable, write "prime." See Examples 1-5. $$ p^{4}-2 p^{3}-8 p+16 $$

Step-by-Step Solution

Answer

The factorization is \((p - 2)^2(p^2 + 2p + 4)\).

Step 1: Identify Factorable Expressions

First, check if the polynomial \(p^4 - 2p^3 - 8p + 16\) can be factored by attempting to recognize patterns or common factor groups. Before grouping, confirm if the expression is in its simplest form.

Step 2: Check for Common Factors

Examine the polynomial to see if there is a common factor across all terms. In this case, each term does not have a common factor other than 1, so proceed with other factoring techniques.

Step 3: Factor by Grouping

Attempt to factor the polynomial by grouping terms. Start by grouping: \[(p^4 - 2p^3) - (8p - 16)\].

Step 4: Factor Each Group

Factor each group separately. From the first group \(p^4 - 2p^3\), factor out \(p^3\) to get \(p^3(p - 2)\). From the second group \(-8p + 16\), factor out \(-8\) to get \(-8(p - 2)\).

Step 5: Factor the Common Binomial

Notice the common binomial factor \((p - 2)\) in each group. The expression is now \(p^3(p - 2) - 8(p - 2)\), which can be rewritten as \((p^3 - 8)(p - 2)\).

Step 6: Recognize Special Products

Recognize that \(p^3 - 8\) is a difference of cubes. Recall that \(a^3 - b^3 = (a - b)(a^2 + ab + b^2)\). Here, \(p^3 - 8\) is \((p)^3 - (2)^3\).

Step 7: Factor the Difference of Cubes

Apply the difference of cubes formula to \(p^3 - 8\): \((p - 2)(p^2 + 2p + 4)\). The complete factorization is now: \[(p - 2)^2(p^2 + 2p + 4)\].

Key Concepts

Algebraic Expressions

In algebra, an expression is a combination of numbers, variables, and mathematical operations. An algebraic expression can be as simple as a single number or as complex as multiple terms. Breaking these down is essential to understand their behavior and how they can be simplified.

When dealing with factoring, the aim is to express the given algebraic expression as a product of its simpler components. For instance, in the expression \( p^4 - 2p^3 - 8p + 16 \), each term is constructed using the variable \( p \) to different powers. These terms are stitched together using addition or subtraction. Simplifying such expressions often involves recognizing these patterns and strategically factoring them.

Factoring in algebra doesn't necessarily simplify the expression numerically but reorganizes it. This can make solving equations easier as it exposes the hidden structure of the equation. It helps in many areas, such as simplifying rational expressions or finding zeros of polynomial functions. Understanding how to manipulate these expressions is a core skill in algebra.

Difference of Cubes

A difference of cubes is a specific pattern where two cubic terms are subtracted. Recognizing this pattern is crucial when factoring expressions like \( p^3 - 8 \). The standard form of a difference of cubes is \( a^3 - b^3 \), where \( a \) and \( b \) are any real numbers or algebraic expressions.

The difference of cubes can be factored using the formula:

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

In our example, \( a = p \) and \( b = 2 \), thus \( p^3 - 8 \) can be rewritten as \( (p - 2)(p^2 + 2p + 4) \). This pattern is efficient for revealing hidden solutions or simplifying polynomials further.

It's important to recognize these special products because they appear quite frequently in algebra problems. Knowing the formulas by heart can save time and effort, as well as lead to more elegant solutions.

Factoring by Grouping

Factoring by grouping is a technique used to simplify polynomials that have four or more terms. The goal is to make the polynomial easier to manage by rearranging and factoring smaller groups of terms. Let's see how to apply this method to \( p^4 - 2p^3 - 8p + 16 \).

First, we split the expression into two pairs: \((p^4 - 2p^3)\) and \(( -8p + 16)\). Next, each group is factored individually:

• For \( p^4 - 2p^3 \), factor out \( p^3 \), resulting in \( p^3(p - 2) \).
• In \(-8p + 16\), factor out \(-8\), leading to \(-8(p - 2) \).

Now, with \( (p - 2) \) appearing in both grouped expressions, it’s identified as a common factor. This allows us to combine them into a single expression: \( (p^3 - 8)(p - 2) \).

Factoring by grouping is especially useful when expressions do not inherently follow a notable form. This approach requires trial and error combined with strategic recognition of common factors to build viable factored forms.