/*! 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 19 Factor completely. $$ 64-125... [FREE SOLUTION] | 91影视

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Chapter 6: Problem 19

Factor completely. $$ 64-125 x^{3} $$

Short Answer

Expert verified
The factored form is \((4 - 5x)(16 + 20x + 25x^2)\).

Step by step solution

01

Identify the Expression Type

Recognize that the given expression, \(64 - 125x^3\), is a difference of cubes. The difference of cubes formula is \(a^3 - b^3 = (a - b)(a^2 + ab + b^2)\).
02

Rewrite as a Difference of Cubes

Rewrite the expression \(64 - 125x^3\) as \((4)^3 - (5x)^3\), so we can clearly see it matches the form \(a^3 - b^3\) where \(a = 4\) and \(b = 5x\).
03

Apply the Difference of Cubes Formula

Using the formula \(a^3 - b^3 = (a - b)(a^2 + ab + b^2)\), substitute \(a = 4\) and \(b = 5x\). This gives us: \((4 - 5x)((4)^2 + 4(5x) + (5x)^2)\).
04

Simplify the Expression

Simplify the terms inside the parenthesis: \((4)^2 = 16\), \(4(5x) = 20x\), and \((5x)^2 = 25x^2\). Therefore, the expression becomes \((4 - 5x)(16 + 20x + 25x^2)\).

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Key Concepts

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

Factoring Polynomials
Factoring polynomials is a key skill in algebra. It involves expressing a polynomial as a product of its factors. To factor a polynomial, you need to recognize patterns and apply mathematical identities.
One common pattern is the difference of cubes, like in the exercise. The steps involve identifying if the polynomial can be written in the form of the difference of cubes, applying the factorization formula, and then simplifying.
This process transforms a complex polynomial into a product of simpler binomials and trinomials, making it easier to solve or simplify further.
Algebraic Expressions
Algebraic expressions use numbers, variables, and operators to represent a value. In the given exercise, the expression is \(64 - 125x^3\).
Understanding what type of algebraic expression you are dealing with is essential. Here, we are working with a polynomial expression, specifically a difference of cubes.
By recognizing \(64 - 125x^3\) as \(4^3 - (5x)^3\), we can apply the difference of cubes formula: \(a^3 - b^3 = (a - b)(a^2 + ab + b^2)\). This transforms the expression into a more workable form. Knowing how to manipulate and work with algebraic expressions is fundamental to mastering algebra.
Polynomial Simplification
Polynomial simplification is reducing a polynomial to its simplest form. In our exercise, it means taking \(64 - 125x^3\) and rewriting it in a factored form using the difference of cubes formula.
This involves breaking it down into smaller, simpler components: \((4 - 5x)(16 + 20x + 25x^2)\). Simplification helps in solving equations and understanding polynomial behavior.
By simplifying, you not only make the polynomial more manageable but also prepare it for potential further algebraic manipulation or solve it for roots.

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Most popular questions from this chapter

To prepare for Section \(6.5,\) review the product and power rules for exponents and multiplication of polynomials (Sections 5.1 and 5.5 ). Multiply.[ 5.5]. $$ (x+1)(x+1)(x+1) $$

Review multiplying binomials using FOIL $$ (x+2)(x-7) $$

Factor completely. Remember to look first for a common factor. If a polynomial is prime, state this. $$ 6 x^{2}-6 y^{2} $$

Ashlee factors \(8 x^{2} y-10 x y^{2}\) as $$ 2 x y \cdot 4 x-2 x y \cdot 5 y $$ Is this the factorization of the polynomial? Why or why not?

To prepare for Section \(6.5,\) review the product and power rules for exponents and multiplication of polynomials (Sections 5.1 and 5.5 ). Simplify. [ 5.1] $$ \left(-5 x^{2} y\right)^{3} $$
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