Problem 51 - Algebra 2

Question

Factor completely. If the polynomial is not factorable, write prime. $$ c^{3}-216 $$

Step-by-Step Solution

Answer

The polynomial is factored as \((c - 6)(c^2 + 6c + 36)\).

Step 1: Recognize the form of the polynomial

The given polynomial is \( c^3 - 216 \). Notice this is a difference of cubes, in the form of \( a^3 - b^3 \), where \( a = c \) and \( b^3 = 216 \), thus \( b = 6 \).

Step 2: Apply the difference of cubes formula

The difference of cubes can be factored using the formula: \( a^3 - b^3 = (a - b)(a^2 + ab + b^2) \). For this polynomial, \( a = c \) and \( b = 6 \). Substitute into the formula to get: \[ c^3 - 216 = (c - 6)(c^2 + 6c + 36) \].

Step 3: Check if the quadratic can be factored further

Examine the quadratic \( c^2 + 6c + 36 \) to see if it can be factored further. Checking the discriminant, \( b^2 - 4ac \) which is \( (6)^2 - 4(1)(36) = 36 - 144 = -108 \), confirms it cannot be factored using real numbers.

Step 4: Write the final factored form

Since \( c^2 + 6c + 36 \) is not factorable over the reals, the completely factored form of the polynomial is \( (c - 6)(c^2 + 6c + 36) \).

Key Concepts

Difference of Cubes

When you see an expression like \( c^3 - 216 \), it's a classic case of a difference of cubes. In algebra, recognizing these forms is crucial for simplifying and factoring them correctly. The general form of a difference of cubes is given by \( a^3 - b^3 \), where both \( a \) and \( b \) are variables or constants raised to the power of three. Here, \( 216 \) is actually \( 6^3 \), making our polynomial \( c^3 - 6^3 \).The formula to factor a difference of cubes is:

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

Substitute \( a = c \) and \( b = 6 \): this gives us:\[ c^3 - 216 = (c - 6)(c^2 + 6c + 36) \]By applying this formula, we simplify the complex expression into a product of simpler terms, which is often the goal when working with polynomials.

Algebraic Expressions

Algebraic expressions are a combination of numbers, variables, and operations that represent a value. In the expression \( c^3 - 216 \), the component \( c^3 \) is a variable term, while \( 216 \) is a constant. The expression itself is a polynomial.Understanding algebraic expressions involves identifying terms, coefficients, and how they are combined. Each term in the expression stands as a building block, which, in this case of factoring, helps recognize it as a difference of cubes.

• The variable \( c \) is raised to the power of three, indicating a cubic term.
• The constant \( 216 \) is significant because it can be rewritten as \( 6^3 \).
• Operations like subtraction (in this case) separate the terms in the expression.

Understanding how expressions are structured can reveal paths to simplification, such as identifying a need to apply specific factoring formulas.

Quadratic Discriminant

The quadratic discriminant is a handy tool in algebra to determine if a quadratic polynomial can be factored over the real numbers. The expression \( c^2 + 6c + 36 \), revealed during the factoring of \( c^3 - 216 \), is a quadratic.The discriminant \( \Delta \) is given by the formula:

• \( \Delta = b^2 - 4ac \)

For the quadratic \( c^2 + 6c + 36 \), use:

• \( b = 6 \) (the coefficient of \( c \))
• \( a = 1 \) (the coefficient of \( c^2 \))
• \( c = 36 \) (the constant term)

Calculate \( \Delta \):\[ \Delta = (6)^2 - 4(1)(36) = 36 - 144 = -108 \]Since \( \Delta \) is negative, this quadratic cannot be factored over the reals, meaning it doesn't have real roots. Therefore, the expression remains as is and part of the complete factorization. Recognizing and computing the discriminant helps decide the factorability of more complex algebraic expressions.