Problem 23 - 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. $$ 14 c-147+c^{2} $$

Step-by-Step Solution

Answer

The expression factors to \((c - 7)(c + 21)\).

Step 1: Identify the Expression

We are given the expression \(c^2 + 14c - 147\). Our goal is to factor this expression.

Step 2: Reorder and Identify Coefficients

Rewrite the quadratic in standard form as \(c^2 + 14c - 147\). Identify the coefficients: \(a = 1\), \(b = 14\), and \(c = -147\).

Step 3: Find Two Numbers That Multiply to the Constant Term

We need two numbers that multiply to \(a \times c = 1 \times (-147) = -147\) and add to \(b = 14\).

Step 4: Determine the Pair That Satisfies Both Conditions

The pair \((21, -7)\) satisfies both conditions: \(21 \times (-7) = -147\) and \(21 + (-7) = 14\).

Step 5: Rewriting the Middle Term

Rewrite the expression as \(c^2 + 21c - 7c - 147\), effectively splitting the middle term.

Step 6: Factor by Grouping

Group the terms: \((c^2 + 21c) + (-7c - 147)\). Factor each group: \(c(c + 21) - 7(c + 21)\).

Step 7: Factor Out the Common Binomial

The expression becomes \((c - 7)(c + 21)\) after factoring out the common factor \((c + 21)\).

Step 8: Verify Factoring

Verify by expanding \((c - 7)(c + 21)\) to ensure it equals the original expression: \((c - 7)(c + 21) = c^2 + 21c - 7c - 147 = c^2 + 14c - 147\).

Key Concepts

Quadratic Equations

Quadratic equations are a central topic in algebra. These are polynomial equations of the form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants, and \( x \) represents an unknown variable. Such equations can produce up to two solutions, due to their degree (which is 2).

Quadratic equations can have real or complex solutions, depending on the discriminant, \( b^2 - 4ac \). A positive discriminant indicates two distinct real roots, a zero discriminant shows one real root (a double root), and a negative discriminant shows two complex roots.

There are multiple ways to solve quadratics, including factoring, completing the square, using the quadratic formula, and graphing. In essence, solving a quadratic means finding the values of \( x \) that make the equation zero. These values are often called "roots" or "solutions" of the quadratic equation. Understanding how to manipulate and solve these equations forms the backbone of many concepts in algebra.

Factoring by Grouping

Factoring by grouping is a method used mainly when dealing with four-term polynomials. However, it is also useful in some trinomials, like the one in our exercise, by splitting the middle term.

To approach factoring by grouping, follow these steps:

• Split the middle term into two terms to create a four-term polynomial that's more manageable.
• Organize the terms into groups, typically two pairs.
• Factor each group separately, looking for common factors.
• Once the common binomial factor is identified in both groups, you can take it out.

In our problem, \( c^2 + 14c - 147 \) was factored by recognizing that splitting the \( 14c \) into \( 21c - 7c \) made it easier to group. Factoring from there allowed the expression to simplify and reveal common factors. This reveals the two binomial factors \((c - 7)(c + 21)\) as the final factorization. Mastering this method will offer insight into solving a wide range of polynomials.

Prime Factorization

Prime factorization involves breaking down a number or expression into its most basic building blocks, known as prime numbers. A prime number is a number greater than 1 that has no positive divisors other than 1 and itself.

When applied to the factorization of expressions such as quadratic polynomials, this concept is leveraged differently. Often, the process involves trial of various combinations, testing factors until the most simplified form is realized, as is the case with factoring by grouping. This process might start with an initial guess based on the coefficients of the polynomial,

For example, in trying to factor \( c^2 + 14c - 147 \), identifying numbers that multiply to a product of \(-147\) and add to \(14\) was crucial. In this scenario, numbers \(21\) and \(-7\) were identified through simple guesses and checks or by computational tools, showing how prime factorization can also inspire solutions in more complex algebraic reductions.

Ultimately, prime factorization helps in breaking down and understanding of the simplest components that can create more complex structures, hence it plays a fundamental role in many areas of mathematics.