/*! 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 171 Factor each polynomial completel... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 5: Problem 171

Factor each polynomial completely. $$z^{3}-125$$

Short Answer

Expert verified
\((z - 5)(z^2 + 5z + 25)\)

Step by step solution

01

Identify the type of polynomial

Recognize that the polynomial is a cubic expression in the form of a difference of cubes: \(z^3 - 125 = z^3 - 5^3\)
02

Apply the difference of cubes formula

Recall the difference of cubes formula: \(a^3 - b^3 = (a - b)(a^2 + ab + b^2)\). Here, \(a = z\) and \(b = 5\).
03

Substitute into the formula

Replace \(a\) and \(b\) with \(z\) and \(5\) respectively: \((z - 5)(z^2 + 5z + 25)\)
04

Verify the factorization

Expand \((z - 5)(z^2 + 5z + 25)\) to ensure it matches the original expression: \((z - 5)(z^2 + 5z + 25) = z^3 + 5z^2 + 25z - 5z^2 - 25z - 125 = z^3 - 125\)

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Difference of Cubes
Recognizing the difference of cubes is crucial when factoring certain polynomials. The difference of cubes formula states that any expression of the form \(a^3 - b^3\) can be factored into \(a - b)(a^2 + ab + b^2)\). This is because when you multiply these two binomials, the terms involving \(a^2b\) and \(ab^2\) cancel out, leaving only the cubic terms and constants. For example, in the polynomial given \(z^3 - 125\), we can see it's written in the form \(a^3 - b^3\) with \(a = z\) and \(b = 5\). So, applying the formula here would give us \(z^3 - 5^3 = (z - 5)(z^2 + 5z + 25)\).
Cubic Expressions
Cubic expressions are polynomials where the highest degree term is a cube, such as \(z^3\). These can often be factored using specific formulas like the difference of cubes. To identify if a polynomial can be factored using the difference of cubes, check if it can be written as \(a^3 - b^3\). For example, a common cubic expression might be \(x^3 - 8\). Here, you can rewrite \(8\) as \(2^3\), yielding \(x^3 - 2^3\). Knowing these forms helps us in reducing complex polynomials into simpler products of binomials and trinomials.
Polynomial Factorization
Polynomial factorization is a method used to break down polynomials into products of simpler polynomials. By factoring polynomials, you simplify the expressions, making them easier to work with, especially for solving equations. For the polynomial \(z^3 - 125\), we recognize it as a special case of a difference of cubes. So, we use the formula \(a^3 - b^3 = (a - b)(a^2 + ab + b^2)\) to factor it into \( (z - 5)(z^2 + 5z + 25) \). Verification is a critical final step. You can confirm this factorization is correct if re-multiplying the factors returns you to the original polynomial: \( (z - 5)(z^2 + 5z + 25) = z^3 - 125 \).

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Factor each polynomial by grouping. $$ y^{2}-a y-b y+a b $$

If you divide each side of \(x^{2}=x\) by \(x,\) you get \(x=1\) If you subtract \(x\) from each side of \(x^{2}=x,\) you get \(x^{2}-x=0,\) which has two solutions. Which method is correct? Explain.

Laying tile. Lorinda is planning to redo the floor in her bedroom, which has an area of 192 square feet. If the width of the rectangular room is 4 feet less than the length, then what are its dimensions?

Factor each polynomial completely. $$ 81 x^{2}-9 $$

Factor each polynomial completely. $$ 9 x^{2}+30 x+25 $$
See all solutions

Recommended explanations on Math Textbooks

Probability and Statistics

Read Explanation

Applied Mathematics

Read Explanation

Mechanics Maths

Read Explanation

Geometry

Read Explanation

Decision Maths

Read Explanation

Pure Maths

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.