/*! 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 126 Factor each polynomial completel... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 5: Problem 126

Factor each polynomial completely. \((a-b)^{3}-(a+b)^{3}\)

Short Answer

Expert verified
The factored form is \[ -2b(3a^{2} + b^{2}) \].

Step by step solution

01

Expand the binomials

First, expand \( (a-b)^{3} \) and \( (a+b)^{3} \) using the binomial theorem. \[(a-b)^{3} = a^{3} - 3a^{2}b + 3ab^{2} - b^{3} \]\[(a+b)^{3} = a^{3} + 3a^{2}b + 3ab^{2} + b^{3} \]
02

Set up the difference

Write the original expression as a difference of the expanded forms:\[(a-b)^{3}-(a+b)^{3} = (a^{3} - 3a^{2}b + 3ab^{2} - b^{3}) - (a^{3} + 3a^{2}b + 3ab^{2} + b^{3}) \]
03

Combine like terms

Combine the like terms by subtracting the expanded forms:\[(a^{3} - 3a^{2}b + 3ab^{2} - b^{3}) - (a^{3} + 3a^{2}b + 3ab^{2} + b^{3}) \]Simplifies to\[a^{3} - 3a^{2}b + 3ab^{2} - b^{3} - a^{3} - 3a^{2}b - 3ab^{2} - b^{3} = -6a^{2}b - 2b^{3} \]
04

Factor out the common terms

Factor out the common terms from the simplified expression:\[ -6a^{2}b - 2b^{3} = -2b(3a^{2} + b^{2}) \]

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.

Understanding the Binomial Theorem
The binomial theorem is an essential tool in algebra for expanding expressions that are raised to a power. It helps you expand polynomials like \( (a-b)^3 \) and \( (a+b)^3 \).

The binomial theorem states that: \ \[ (x + y)^n = \binom{n}{0} x^n y^0 + \binom{n}{1} x^{n-1} y^1 + ... + \binom{n}{n} x^0 y^n \] where \( \binom{n}{k} \) is a binomial coefficient.

For example, to expand \( (a-b)^3 \) using the binomial theorem: \[ (a-b)^3 = a^3 - 3a^2b + 3ab^2 - b^3 \] and \( (a+b)^3 \) expands to: \[ (a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3 \]
Understanding this expansion will help you in further steps of polynomial factorization.
Difference of Cubes
The difference of cubes formula is another useful tool in polynomial factorization.

It's used to simplify expressions of the form \( a^3 - b^3 \). The formula for the difference of cubes is: \[ a^3 - b^3 = (a - b)(a^2 + ab + b^2) \] While this problem isn't directly about difference of cubes, it involves a similar pattern once the binomials are expanded and combined. Recognizing patterns like these is key to mastering polynomial factorization.

Try applying this concept step by step to various polynomial problems to get more familiar with it.
Factoring Polynomials
Factoring polynomials involves breaking down a polynomial into simpler 'factor' polynomials that, when multiplied together, give back the original polynomial.

Using our exercise example, after expanding and combining terms, we reached the expression \( -6a^2b - 2b^3 \).
The next step is to factor out the greatest common factor, which in this case is \( -2b \): \[ -6a^2b - 2b^3 = -2b(3a^2 + b^2) \]

Factoring polynomials might involve recognizing special patterns (like the difference of cubes) or simply factoring out common terms. Practice identifying these patterns and common factors to become more comfortable with polynomial factorization.

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

The following powers of \(x\) are all perfect cubes: $$ x^{3}, x^{6}, \quad x^{9}, \quad x^{12}, \quad x^{15} $$ On the basis of this observation, we may make a conjecture that if the power of a variable is divisible by ____ (with 0 remainder), then we have a perfect cube.

Factor each trinomial completely. \(6 m^{6} n+7 m^{5} n^{2}+2 m^{4} n^{3}\)

Solve each problem. Tram works due north of home. Her husband Alan works due east. They leave for work at the same time. By the time Tram is \(5 \mathrm{mi}\) from home, the distance between them is \(1 \mathrm{mi}\) more than Alan's distance from home. How far from home is Alan?

Factor each trinomial. \(24 x^{4}+17 x^{2}-20\)

Solve each problem. The product of two consecutive integers is 11 more than their sum. Find the integers.
See all solutions

Recommended explanations on Math Textbooks

Probability and Statistics

Read Explanation

Statistics

Read Explanation

Decision Maths

Read Explanation

Mechanics Maths

Read Explanation

Logic and Functions

Read Explanation

Applied Mathematics

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.