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Chapter 6: Problem 78

Factor completely. Identify any prime polynomials. $$ 2 a^{2}-32 b^{2} $$

Short Answer

Expert verified
The factored form is \(2(a + 4b)(a - 4b)\).

Step by step solution

01

Identify the common factor

First, look for any common factors between the terms. In the polynomial \(2a^2 - 32b^2\), the common factor is 2. Factor out the 2 to get: \[ 2(a^2 - 16b^2) \]
02

Recognize the difference of squares

The expression inside the parentheses, \(a^2 - 16b^2\), is a difference of squares because it can be written as \(a^2 - (4b)^2\). Recall that the difference of squares formula is \(x^2 - y^2 = (x + y)(x - y)\).
03

Apply the difference of squares formula

Apply the difference of squares formula to the expression \(a^2 - 16b^2\): \[a^2 - (4b)^2 = (a + 4b)(a - 4b) \]
04

Combine the factors

Multiply the factored terms by the common factor from Step 1: \[ 2(a + 4b)(a - 4b) \]
05

Identify prime polynomials

Check each factor to see if it can be factored further. In this case, \(a + 4b\) and \(a - 4b\) are both prime polynomials.

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Key Concepts

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

Common Factors
In algebra, common factors are essential when simplifying and factoring polynomials. A common factor is a number or variable that divides two or more terms evenly.
In our example, the polynomial given is \(2a^2 - 32b^2\).
To factor it, start by identifying any common factors in each term. Here, the common factor between \(2a^2\) and \(32b^2\) is 2. This means that both terms can be divided by 2.
When we factor out the common factor of 2, we get: \[ 2(a^2 - 16b^2) \].
Recognizing common factors can make polynomials easier to work with and is a crucial first step in the process of factoring.
Difference of Squares
The difference of squares is a particular type of polynomial that can be factored using a specific formula. It usually takes the form \(x^2 - y^2\).
You can recognize it because it features the subtraction of one squared term from another. In our polynomial \(a^2 - 16b^2\), notice that both \(a^2\) and \(16b^2\) are perfect squares (since \(16b^2 = (4b)^2\)). This leads to: \(a^2 - (4b)^2\).
Applying the difference of squares formula gives us: \[ x^2 - y^2 = (x + y)(x - y) \].
So, for our example: \[ a^2 - (4b)^2 = (a + 4b)(a - 4b) \].
This method streamlines the process of factoring certain polynomials significantly, making it easier to break them down into simpler factors.
Prime Polynomials
Prime polynomials are the polynomials that cannot be factored any further into products of polynomials with integer coefficients. Think of them as the polynomials equivalent of prime numbers.
In our solution, after factoring \(a^2 - 16b^2\) using the difference of squares, we end up with the factors \((a + 4b)\) and \((a - 4b)\).
The next step is to check if these factors can be broken down further. After testing, we find that neither \((a + 4b)\) nor \((a - 4b)\) can be factored further. Therefore, these polynomials are considered prime.
Recognizing prime polynomials is useful because it tells us when we have reached the simplest form of our factored polynomial.

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Most popular questions from this chapter

The width of a rectangle is \(13 \mathrm{ft}\) less than twice its length. Its area is \(45 \mathrm{ft}^{2}\).

(a) Describe the mistake in words, or copy down the whole problem and highlight or circle the mistake. (b) Do the problem correctly. Problem: Factor \(45 c^{3}+33 c^{2}-168 c\) completely. Incorrect Answer: $$ \begin{aligned} &45 c^{3}+33 c^{2}-168 c \\ &=3 c\left(15 c^{2}+11 c-56\right) \end{aligned} $$

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Factor completely. Identify any prime polynomials. $$ p^{3}+27 z^{3} $$
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