Problem 62
Question
Factor the expression completely. $$ x^{3}-27 $$
Step-by-Step Solution
Verified Answer
The expression \( x^3 - 27 \) factors to \((x - 3)(x^2 + 3x + 9)\).
1Step 1: Identify the expression as a difference of cubes
The expression given is \( x^3 - 27 \). Recognize this as a difference of cubes. We know that both \( x^3 \) and \( 27 \) are perfect cubes: \( x^3 = (x)^3 \) and \( 27 = (3)^3 \). Thus, \( x^3 - 27 \) is in the form \( a^3 - b^3 \) where \( a = x \) and \( b = 3 \).
2Step 2: Use the difference of cubes formula
The difference of cubes formula is \( a^3 - b^3 = (a - b)(a^2 + ab + b^2) \). Substitute \( a = x \) and \( b = 3 \) into the formula to factor the expression. The factored form becomes \((x - 3)(x^2 + 3x + 9)\).
3Step 3: Verify the factorization
To ensure the factorization is correct, expand the factored expression \((x - 3)(x^2 + 3x + 9)\). This results in: \[ x(x^2) + x(3x) + x(9) - 3(x^2) - 3(3x) - 3(9) \] which simplifies to \( x^3 + 3x^2 + 9x - 3x^2 - 9x - 27 \), and then to \( x^3 - 27 \). The expanded form matches the original expression, confirming the factorization.
Key Concepts
Factoring PolynomialsAlgebraic ExpressionsPolynomial Identities
Factoring Polynomials
Factoring polynomials is a technique used to simplify algebraic expressions by breaking them down into multiple factors. Just like numbers can be expressed as a product of their prime factors, polynomials can be factored into simpler components. For example, consider the polynomial expression you encountered: \( x^3 - 27 \). This is a special kind of polynomial called a difference of cubes. Here's how you can factor such expressions:
- Identify the terms as perfect cubes. For instance, \( x^3 \) and \( 27 \) are both perfect cubes.
- Recognize the form \( a^3 - b^3 \) and apply the appropriate identity.
- Use the formula \( a^3 - b^3 = (a - b)(a^2 + ab + b^2) \) to factor.
Algebraic Expressions
Algebraic expressions are combinations of numbers, variables, and operations. They can represent real-world situations and are the foundation of algebra. Understanding them creates a strong base for solving equations and other mathematical problems. The expression \( x^3 - 27 \) is an algebraic expression that combines both a variable (\( x \)) and a constant (27). Key components you may encounter in algebraic expressions include:
- Coefficients: The numerical part that multiplies a variable.
- Variables: Symbols like \( x \) that represent unknown values.
- Constants: Fixed numbers like 27.
- Operators: Such as addition, subtraction, or multiplication.
Polynomial Identities
Polynomial identities are equations that are true for every value of the variable in them. These identities allow mathematicians to simplify polynomial expressions and solve equations efficiently.The difference of cubes is a specific polynomial identity that simplifies expressions like \( x^3 - 27 \). The identity is given by:\[a^3 - b^3 = (a - b)(a^2 + ab + b^2)\]This identity states that any difference of two cubes can be broken down into a product of a linear factor \((a-b)\) and a quadratic factor \((a^2 + ab + b^2)\).Understanding and applying polynomial identities:
- Helps in breaking down complex expressions into simpler parts.
- Allows for easier calculations and manipulations.
- Provides insight into the nature and behavior of polynomial expressions.
Other exercises in this chapter
Problem 62
Perform the addition or subtraction and simplify. $$ \frac{x}{x^{2}-x-6}-\frac{1}{x+2}-\frac{2}{x-3} $$
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\(47-72\) . Simplify the expression, and eliminate any negative exponent(s). $$ \left(\frac{x^{4} z^{2}}{4 y^{5}}\right)\left(\frac{2 x^{3} y^{2}}{z^{3}}\right)
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Find the distance between the given numbers. $$ \begin{array}{llll}{\text { (a) } \frac{7}{15} \text { and }-\frac{1}{21}} & {\text { (b) }-38 \text { and }-57}
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\(49-68=\) Simplify the expression and eliminate any negative exponent(s). Assume that all letters denote positive numbers. $$ \frac{\left(32 y^{-5} z^{10}\righ
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