/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 166 Factor each polynomial completel... [FREE SOLUTION] | 91影视

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Chapter 5: Problem 166

Factor each polynomial completely. $$a^{3}-64$$

Short Answer

Expert verified
The complete factorization is \((a - 4)(a^{2} + 4a + 16)\).

Step by step solution

01

Identify the form of the polynomial

The given polynomial is \(a^{3}-64\). Recognize that it is a difference of cubes. Generally, a difference of cubes can be expressed as \(a^{3} - b^{3}\). Here, \(a^{3}\) is already given and \(64\) can be written as \(4^{3}\).
02

Apply the difference of cubes formula

The difference of cubes formula is \(a^{3} - b^{3} = (a - b)(a^{2} + ab + b^{2})\). Substitute \(a\) and \(b\) with the terms identified in step 1: \(a\) is \(a\) and \(b\) is \(4\). Therefore, \(a^{3} - 4^{3}\) becomes \((a - 4)(a^{2} + 4a + 16)\).
03

Verify the factorization

Verify the factorization by expanding \((a - 4)(a^{2} + 4a + 16)\) to check if it gives back the original polynomial. Expand \((a - 4)\): \(a \cdot a^{2} + a \cdot 4a + a \cdot 16 - 4 \cdot a^{2} - 4 \cdot 4a - 4 \cdot 16\). Simplify the terms: \(a^{3} + 4a^{2} + 16a - 4a^{2} - 16a - 64\). Combine like terms to verify it results in \(a^{3} - 64\).

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Key Concepts

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

Difference of Cubes
The **difference of cubes** is a special form of polynomial represented as \(a^3 - b^3\). This format allows for easy factorization using a specific algebraic identity. In our example, the polynomial is given as \(a^3 - 64\). Recognize that \(64\) is a perfect cube, as it can be written as \(4^3\). So, the given expression transforms into \(a^3 - 4^3\), making it fit the **difference of cubes** format perfectly. Here, \(a = a\) and \(b = 4\). The formula for factorizing the difference of cubes is:
Factorization
**Factorization** involves breaking down a complex expression into simpler parts (or factors) that, when multiplied together, give back the original expression. In the example \(a^3 - 64\), we performed factorization by identifying it as a difference of cubes. The process typically involves:
  • Recognizing the polynomial's form (in this case, \(a^3 - b^3\))
  • Applying the appropriate algebraic identity
  • Simplifying the resulting factors
For \(a^3 - 64\), after recognizing it as \(a^3 - 4^3\), we applied the difference of cubes formula:
Algebraic Identities
**Algebraic identities** are essential tools that provide simplified formulas for common polynomial forms. They make factorization and expansion much easier. Some commonly used identities are:
  • Difference of squares: \(a^2 - b^2 = (a - b)(a + b)\)
  • Sum of cubes: \(a^3 + b^3 = (a + b)(a^2 - ab + b^2)\)
  • Difference of cubes: \(a^3 - b^3 = (a - b)(a^2 + ab + b^2)\)
In our specific case, we used the difference of cubes identity to factorize \(a^3 - 64\). Identifying the appropriate algebraic identity is crucial as it simplifies the polynomial to its factors almost instantly. These identities save time and help in solving polynomials efficiently.

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