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Chapter 6: Problem 13

Factor completely. Identify any prime polynomials. $$ 18 a^{2} c+42 a^{2}+45 a c k+105 a k $$

Short Answer

Expert verified
The factored form is \(3a (3c + 7) (2a + 5k)\).

Step by step solution

01

Factor out the Greatest Common Factor (GCF)

Identify the GCF of all terms in the expression. The terms are: 1. \(18 a^2 c\), 2. \(42 a^2\), 3. \(45 a c k\), 4. \(105 a k\). The GCF here is \(3a\). Factor \(3a\) out of each term: \[ 3a (6ac + 14a + 15ck + 35k). \]
02

Group the terms

Group the terms in pairs to factor by grouping. Group: \[ 3a [(6ac + 14a) + (15ck + 35k)]. \]
03

Factor out the Greatest Common Factor in each group

Factor out the GCF from each grouped term: \[ 3a [2a (3c + 7) + 5k (3c + 7)]. \]
04

Identify and factor out the common binomial

Notice the common binomial \((3c + 7)\) in both terms. Factor it out: \[ 3a (3c + 7) (2a + 5k). \]
05

Confirm factored form and identify any prime polynomials

The final factored form is: \[ 3a (3c + 7) (2a + 5k). \] None of the factors \((3a, 3c + 7, 2a + 5k)\) are prime polynomials.

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Key Concepts

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

Greatest Common Factor
One of the most useful methods for factoring polynomials is finding the Greatest Common Factor (GCF). The GCF is the largest factor that can evenly divide each term in the polynomial. In the given exercise, the terms are \(18a^2c\), \(42a^2\), \(45ack\), and \(105ak\). The GCF of these terms is \(3a\).

By factoring out \(3a\) from each term, we simplify the expression. This reduced form is the first step towards further factoring. In our case, factoring out \(3a\) gives us:
  • \( 3a(6ac + 14a + 15ck + 35k) \).
This makes the subsequent steps easier.
factoring by grouping
Factoring by grouping is a technique used for polynomials with four or more terms. The goal is to group terms with common factors and then factor out these common factors. After we factor out \(3a\), we have \(3a(6ac + 14a + 15ck + 35k)\).

Grouping the terms in pairs, we get:
  • \(3a [ (6ac + 14a) + (15ck + 35k)]\).
Now, we can factor out the GCF from each group. For the first group, \((6ac + 14a)\), the GCF is \(2a\); and for the second group, \((15ck + 35k)\), the GCF is \(5k\). This gives us:
  • \(3a [ 2a(3c + 7) + 5k(3c + 7) ]\).
Here, you can see that \((3c + 7)\) is a common binomial.
prime polynomials
Prime polynomials are polynomials that cannot be factored further. They are the polynomial equivalent of prime numbers. In our given exercise, after factoring out the GCF and using factoring by grouping, we get the expression:
  • \(3a(3c + 7)(2a + 5k)\).
To confirm that these factors cannot be factored further, we must check if any of them are prime.
  • \(3a\) is already simplified.
  • \( (3c + 7) \) has no common factors and cannot be factored further, making it a prime polynomial.
  • \((2a + 5k)\) also has no common factors and thus is a prime polynomial.
This means the original polynomial is fully factored into its prime components.

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Most popular questions from this chapter

(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 \(6 x^{2}+21 x+15\) completely. $$ \text { Incorrect Answer: } \begin{aligned} & 6 x^{2}+21 x+15 \\ &=3\left(2 x^{2}+7 x+5\right) \end{aligned} $$

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