Is (x+2) A Factor Of 3x^2-x-14 Exploring Polynomial Relationships
Deciphering the relationship between a linear expression and a polynomial is a fundamental skill in algebra. In this article, we will delve into the specifics of determining whether is a factor of the polynomial . We'll explore different methods, including the factor theorem and polynomial division, to arrive at a definitive answer. Our journey will not only provide the solution but also enhance your understanding of polynomial factorization and its applications. So, let's embark on this mathematical exploration and uncover the connection between and the given polynomial.
Methods to Determine if (x+2) is a Factor
The Factor Theorem: A Powerful Tool
One of the most efficient ways to determine if is a factor of the polynomial is by employing the factor theorem. This theorem states that for a polynomial , is a factor if and only if . In our case, we want to check if is a factor, which can be rewritten as . Therefore, we need to evaluate the polynomial at .
Let's substitute into the polynomial:
Since , according to the factor theorem, is indeed a factor of the polynomial . This method provides a quick and direct way to ascertain the factor relationship without the need for more complex procedures like polynomial division. The factor theorem is a cornerstone in polynomial algebra, simplifying the process of factorization and root finding.
Polynomial Division: A Comprehensive Approach
Another method to determine if is a factor of is through polynomial division. This method not only tells us whether is a factor but also provides the quotient polynomial if it is. If the division results in a remainder of zero, then is a factor. Let's perform the polynomial division:
Divide by :
3x - 7
x + 2 | 3x^2 - x - 14
- (3x^2 + 6x)
---------
-7x - 14
- (-7x - 14)
---------
0
The division process unfolds as follows: First, we divide the leading term by , resulting in . We then multiply by to get and subtract it from the original polynomial. This yields . Next, we divide by , obtaining . Multiplying by gives , and subtracting this from results in a remainder of .
Since the remainder is , this confirms that is a factor of . Furthermore, the quotient we obtained, , is the other factor of the polynomial. Thus, we can express the polynomial as a product of its factors: . This method not only verifies the factor relationship but also provides a complete factorization of the polynomial.
Factoring by Grouping: An Alternative Technique
While the factor theorem and polynomial division are effective methods, we can also explore factoring by grouping to determine if is a factor of the polynomial . This technique involves rewriting the middle term of the quadratic expression in such a way that we can group terms and factor out common factors. The goal is to express the polynomial as a product of two binomials, one of which should be if it is indeed a factor.
Let's rewrite the polynomial:
We need to find two numbers that multiply to and add up to (the coefficient of the term). These numbers are and . So, we can rewrite the middle term as :
Now, we group the terms:
Factor out the greatest common factor (GCF) from each group:
Notice that is a common factor in both terms. We can factor it out:
By factoring by grouping, we have successfully expressed the polynomial as a product of and . This confirms that is a factor of . This method provides an alternative approach to factorization and reinforces the understanding of how factors relate to the polynomial expression.
Conclusion: (x+2) is Indeed a Factor
Through our exploration using the factor theorem, polynomial division, and factoring by grouping, we have conclusively determined that is indeed a factor of the polynomial . The factor theorem provided a swift verification by showing that . Polynomial division gave us a comprehensive view, confirming the factor relationship and revealing the other factor, . Factoring by grouping offered an alternative pathway to factorization, reinforcing our result. Understanding these methods not only helps in solving this specific problem but also equips us with essential tools for tackling a wide range of polynomial factorization challenges. The interplay between factors and polynomials is a cornerstone of algebraic manipulation, and mastering these techniques opens doors to more advanced mathematical concepts.
Therefore, the correct answer is B: is a factor.