Factor each polynomial. See Examples 5 and 6. 125x3-27

Ch. R - Review of Basic Concepts

Lial13th EditionCollege AlgebraISBN: 9780136881063Not the one you use?Change textbook

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Lial 13th Edition

Ch. R - Review of Basic Concepts

Problem 75

Chapter 1, Problem 75

Factor each polynomial. See Examples 5 and 6. 125x³-27

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Recognize that the polynomial $125x^3 - 27$ is a difference of cubes because $125x^3 = (5x)^3$ and $27 = 3^3$.

Recall the difference of cubes factoring formula: $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$.

Identify $a = 5x$ and $b = 3$ in the expression $125x^3 - 27$.

Apply the formula by substituting $a$ and $b$: write the factorization as $(5x - 3)((5x)^2 + (5x)(3) + 3^2)$.

Simplify the terms inside the second factor: $(5x)^2 = 25x^2$, $(5x)(3) = 15x$, and $3^2 = 9$, so the factorization becomes $(5x - 3)(25x^2 + 15x + 9)$.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Difference of Cubes

The difference of cubes formula is used to factor expressions of the form a³ - b³. It states that a³ - b³ = (a - b)(a² + ab + b²). Recognizing this pattern allows you to factor certain cubic polynomials efficiently.

Identifying Perfect Cubes

To apply the difference of cubes formula, you must identify terms that are perfect cubes. For example, 125x³ is (5x)³ and 27 is 3³. Recognizing these helps rewrite the polynomial in the form a³ - b³.

Polynomial Factoring Techniques

Factoring polynomials involves rewriting them as products of simpler polynomials. Understanding various factoring methods, such as factoring out the greatest common factor or special formulas like difference of cubes, is essential for simplifying expressions.

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