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Chapter 6: Problem 49

(a) factor by grouping. Identify any prime polynomials. (b) check. $$ 3 m^{2}+3 m v+m+v $$

Short Answer

Expert verified
The factored form is \( (m + v)(3m + 1) \). Both binomials are prime.

Step by step solution

01

- Group the terms

Group the polynomial into two pairs: \[ (3m^2 + 3mv) + (m + v) \]
02

- Factor out the greatest common factor (GCF) from each group

Factor out the GCF from each pair:\[ 3m(m + v) + 1(m + v) \]
03

- Factor out the common binomial

Since both terms contain \( (m + v) \), factor \( (m + v) \) out:\[ (m + v)(3m + 1) \]
04

- Identify prime polynomial if necessary

No need to factor further as the expression \( (m + v)(3m + 1) \) cannot be factored further. Both binomials are prime.
05

- Verify by expanding

Expand the factored form to verify correctness:\[ (m + v)(3m + 1) = m(3m + 1) + v(3m + 1) = 3m^2 + m + 3mv + v = 3m^2 + 3mv + m + v \]

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Key Concepts

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

Prime Polynomials
A polynomial is considered 'prime' if it cannot be factored into simpler polynomials over the set of integers. In other words, it's like the polynomial equivalent of a prime number. In the context of our exercise, we found the completely factored form of the polynomial as \((m + v)(3m + 1)\).
To check if these binomials are prime, we try to factor them further.
For \(m + v\), there are no integer factors other than 1 and itself. For \(3m + 1\), neither does it have any factors. Hence, these binomials are prime polynomials.
Remember: Identifying a prime polynomial ensures we've simplified as far as possible, and there are no additional factors hiding in the expression.
Greatest Common Factor (GCF)
The greatest common factor (GCF) is the highest number that can divide each term in a polynomial without leaving a remainder. It's like finding the biggest piece that fits into all terms.
In our exercise, we grouped the polynomials into two pairs: \((3m^2 + 3mv) + (m + v)\).
The GCF of the first pair \(3m^2 + 3mv\) is 3m, and the GCF of the second pair \(m + v\) is 1 since it鈥檚 already in its simplest form.
So, we factor out these GCFs and rewrite the pairs as:
  • First pair: \3m(m + v)\
  • Second pair: \1(m + v)\

    • This step simplifies our work, allowing us to factor out the common binomial \(m + v\), leading to the completely factored form \((m + v)(3m + 1)\). Always look for the GCF first when factoring polynomials. It'll make your work much easier!
Binomials
Binomials are polynomials with exactly two terms. They're like a pair in the world of polynomials.
In our exercise, we arrived at two binomials: \((m + v)\) and \((3m + 1)\). Each binomial has two terms.
  • \((m + v)\) has the terms: m and v.
  • \((3m + 1)\) has the terms: 3m and 1.

    • Working with binomials often involves factoring them, combining like terms, or using them in further operations.
    Binomials can be as simple or as tricky as any polynomial. But in our problem, once we've factored by grouping and identified the common binomial, we could easily see the solution. Remember: Recognizing and naming binomials correctly is fundamental in polynomial algebra.

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