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Chapter 5: Problem 37

Factor each polynomial. $$ x^{3}-27 $$

Short Answer

Expert verified
The factorized form of $$x^3 - 27$$ is $$(x - 3)(x^2 + 3x + 9)$$.

Step by step solution

01

Identify the Special Form

Recognize that the given polynomial, \textbf{ $$x^3 - 27$$ }, is in the form of a difference of cubes: \textbf{ $$a^3 - b^3$$ }, where \textbf{ \(a = x\) } and \textbf{ \(b = 3\)
02

Recall the Difference of Cubes Formula

The formula for factoring a difference of cubes is: \textbf{ $$a^3 - b^3 = (a - b)(a^2 + ab + b^2)$$ }. Use this formula to factor the polynomial.
03

Substitute and Simplify

Substitute \textbf{ \(a = x\) } and \textbf{ \(b = 3\) } into the formula: $$x^3 - 27 = (x - 3)(x^2 + (x)(3) + 3^2)$$
04

Final Simplification

Simplify the polynomial within the parentheses: $$x^3 - 27 = (x - 3)(x^2 + 3x + 9)$$

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Difference of Cubes
A polynomial in the form of a difference of cubes appears like this: \ a^3 - b^3. This means we have two terms that are each raised to the power of three and are subtracted from one another. In the given exercise, $$ x^{3}-27 $$ is identified as a difference of cubes. Here, \( a \) is \( x \) and \( b \) is \( 3 \). Now, remember the difference of cubes formula: \ $$a^3 - b^3 = (a - b)(a^2 + ab + b^2)$$. Using this formula, we can factor the polynomial efficiently.
Polynomial Factorization
Factoring polynomials involves breaking them down into simpler expressions that can be multiplied together to get the original polynomial. This technique is crucial in algebra to simplify expressions and solve equations. By identifying special forms like the difference of cubes, we can apply specific formulas to factorize them quickly. For instance, in our exercise, we substitute \( a = x \) and \( b = 3 \) into the difference of cubes formula, and simplify accordingly: \ $$x^3 - 27 = (x - 3)(x^2 + 3x + 9)$$. This factorization allows us to work with simpler linear and quadratic forms, aiding in further calculations and graphing.
Algebraic Identities
Algebraic identities are equations that hold true for any values of the variables involved. They are powerful tools for simplifying and solving algebraic expressions. One useful identity is the difference of cubes: $$a^3 - b^3 = (a - b)(a^2 + ab + b^2)$$. In our exercise, applying this identity helps us transform the polynomial \( x^3 - 27 \) quickly: \ First, we identify \( x^3 \) as \( a^3 \) and \( 27 \) as \( 3^3 \), then apply the identity to get: \ $$x^3 - 27 = (x - 3)(x^2 + 3x + 9)$$. Understanding and using algebraic identities makes algebra simpler and more manageable.

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