Which Polynomial Has (3x + 2) As A Factor? Step-by-Step Solutions

When faced with the question, "Which polynomial has (3x + 2) as a binomial factor?", it's crucial to understand the underlying concepts of polynomial factorization and the factor theorem. This article provides a detailed explanation of how to solve this type of problem, offering step-by-step solutions and insights into the mathematical principles involved. We will explore different methods to determine if a given binomial is a factor of a polynomial, ensuring a comprehensive understanding for students and math enthusiasts alike.

Understanding Polynomial Factors

In the realm of algebra, polynomials are expressions consisting of variables and coefficients, combined using addition, subtraction, and multiplication. Factoring a polynomial involves breaking it down into simpler expressions (factors) that, when multiplied together, yield the original polynomial. A binomial factor is a polynomial with two terms. Identifying binomial factors is a fundamental skill in algebra, essential for solving equations, simplifying expressions, and understanding the behavior of polynomial functions.

The factor theorem plays a pivotal role in determining if a binomial (ax + b) is a factor of a polynomial P(x). The factor theorem states that (ax + b) is a factor of P(x) if and only if P(-b/a) = 0. This theorem provides a straightforward method for checking potential factors by substituting the root of the binomial into the polynomial. For our specific problem, the binomial factor is (3x + 2), so we need to find the root of this binomial and substitute it into each of the given polynomials.

Method 1: The Factor Theorem

The factor theorem is our primary tool for solving this problem. It allows us to quickly check if $(3x + 2)$ is a factor of a given polynomial. First, we need to find the root of the binomial $(3x + 2)$.

1. Find the root of the binomial: Set $3x + 2 = 0$ and solve for $x$: $3x = -2$ $x = -\frac{2}{3}$

2. Substitute the root into each polynomial: We will now substitute $x = -\frac{2}{3}$ into each of the given polynomials (A, B, C, and D) and check if the result is 0. If the result is 0, then $(3x + 2)$ is a factor of that polynomial.

Option A: $6x^3 + 3x^2 + 4x + 2$

Substitute $x = -\frac{2}{3}$ into the polynomial:

$6(-\frac{2}{3})^3 + 3(-\frac{2}{3})^2 + 4(-\frac{2}{3}) + 2$

$= 6(-\frac{8}{27}) + 3(\frac{4}{9}) - \frac{8}{3} + 2$

$= -\frac{16}{9} + \frac{4}{3} - \frac{8}{3} + 2$

To add these fractions, we need a common denominator, which is 9:

$= -\frac{16}{9} + \frac{12}{9} - \frac{24}{9} + \frac{18}{9}$

$= \frac{-16 + 12 - 24 + 18}{9}$

$= \frac{-10}{9}$

Since the result is not 0, $(3x + 2)$ is not a factor of $6x^3 + 3x^2 + 4x + 2$.

Option B: $12x^3 + 15x^2 + 8x + 10$

Substitute $x = -\frac{2}{3}$ into the polynomial:

$12(-\frac{2}{3})^3 + 15(-\frac{2}{3})^2 + 8(-\frac{2}{3}) + 10$

$= 12(-\frac{8}{27}) + 15(\frac{4}{9}) - \frac{16}{3} + 10$

$= -\frac{32}{9} + \frac{20}{3} - \frac{16}{3} + 10$

To add these terms, we need a common denominator, which is 9:

$= -\frac{32}{9} + \frac{60}{9} - \frac{48}{9} + \frac{90}{9}$

$= \frac{-32 + 60 - 48 + 90}{9}$

$= \frac{70}{9}$

Since the result is not 0, $(3x + 2)$ is not a factor of $12x^3 + 15x^2 + 8x + 10$.

Option C: $18x^3 - 12x^2 + 9x - 6$

Substitute $x = -\frac{2}{3}$ into the polynomial:

$18(-\frac{2}{3})^3 - 12(-\frac{2}{3})^2 + 9(-\frac{2}{3}) - 6$

$= 18(-\frac{8}{27}) - 12(\frac{4}{9}) - 6 - 6$

$= -\frac{16}{3} - \frac{16}{3} - 6 - 6$

$= -\frac{32}{3} - 12$

$= -\frac{32}{3} - \frac{36}{3}$

$= -\frac{68}{3}$

Since the result is not 0, $(3x + 2)$ is not a factor of $18x^3 - 12x^2 + 9x - 6$.

Option D: $21x^4 + 14x^3 + 6x + 4$

Substitute $x = -\frac{2}{3}$ into the polynomial:

$21(-\frac{2}{3})^4 + 14(-\frac{2}{3})^3 + 6(-\frac{2}{3}) + 4$

$= 21(\frac{16}{81}) + 14(-\frac{8}{27}) - 4 + 4$

$= \frac{112}{27} - \frac{112}{27} - 4 + 4$

$= 0$

Since the result is 0, $(3x + 2)$ is a factor of $21x^4 + 14x^3 + 6x + 4$.

Detailed Analysis of the Correct Polynomial

The correct answer is Option D: $21x^4 + 14x^3 + 6x + 4$. This is because when we substituted $x = -\frac{2}{3}$ into the polynomial, the result was 0. According to the factor theorem, this confirms that $(3x + 2)$ is indeed a factor of this polynomial. Let's delve deeper into why this polynomial satisfies the condition while the others do not.

The factor theorem provides a definitive test. However, it's also instructive to understand how factorization might work conceptually. If $(3x + 2)$ is a factor, it means that the polynomial can be written in the form $(3x + 2) imes Q(x)$, where $Q(x)$ is another polynomial. In Option D, we found that plugging in $x = -\frac{2}{3}$ made the polynomial equal to zero, satisfying this condition.

To further illustrate, we could perform polynomial long division or synthetic division to divide $21x^4 + 14x^3 + 6x + 4$ by $(3x + 2)$. This process would yield another polynomial $Q(x)$ without any remainder, confirming that $(3x + 2)$ is a factor. Although not required for this problem, this step provides a more comprehensive understanding of polynomial factorization.

Common Pitfalls and How to Avoid Them

When solving problems involving the factor theorem, several common errors can occur. Being aware of these pitfalls can help you avoid them and improve your accuracy.

One common mistake is incorrect substitution. Ensure that the root of the binomial is calculated correctly and substituted accurately into the polynomial. Double-check your arithmetic when evaluating the polynomial at the substituted value. Even a small error in calculation can lead to an incorrect conclusion.

Another pitfall is misinterpreting the factor theorem. Remember, $(ax + b)$ is a factor of $P(x)$ if and only if $P(-\frac{b}{a}) = 0$. A non-zero result indicates that the binomial is not a factor. Make sure to clearly understand this condition.

Lastly, rushing through the problem can lead to mistakes. Take your time, especially when dealing with fractions and negative signs. A careful, step-by-step approach will minimize errors and increase your confidence in your solution.

Additional Methods for Checking Factors

While the factor theorem is the most efficient method for this particular problem, it's beneficial to be aware of other techniques for checking factors of polynomials. These methods can provide alternative approaches and enhance your overall understanding of polynomial factorization.

Polynomial Long Division

Polynomial long division is a method for dividing polynomials similar to long division with numbers. If dividing a polynomial $P(x)$ by $(3x + 2)$ results in a remainder of 0, then $(3x + 2)$ is a factor of $P(x)$. This method is more time-consuming than the factor theorem but can be useful in certain situations, especially when you need to find the other factor $Q(x)$.

Synthetic Division

Synthetic division is a shortcut method for dividing a polynomial by a linear divisor (of the form $x - c$). It's a more streamlined process than long division and is particularly efficient when the divisor is a simple binomial. If the remainder is 0, then the divisor is a factor of the polynomial.

Factoring by Grouping

Factoring by grouping is a technique used when a polynomial can be divided into groups of terms that share a common factor. While not directly applicable to all polynomials, it can be a useful strategy when dealing with expressions that have four or more terms. This method involves grouping terms, factoring out common factors from each group, and then factoring out a common binomial factor.

Real-World Applications of Polynomial Factorization

Polynomial factorization is not just an abstract mathematical concept; it has numerous real-world applications in various fields. Understanding polynomial factors is essential in engineering, physics, computer science, and economics. Here are a few examples:

Engineering

In engineering, polynomials are used to model various physical systems, such as electrical circuits, mechanical systems, and fluid dynamics. Factoring polynomials helps engineers analyze the stability and behavior of these systems. For example, the roots of a characteristic polynomial can determine the stability of a control system.

Physics

In physics, polynomials appear in equations describing motion, energy, and other physical quantities. Factoring these polynomials can help solve problems related to projectile motion, oscillations, and wave phenomena. The factor theorem and polynomial division can be used to find solutions to complex physical equations.

Computer Science

In computer science, polynomials are used in cryptography, coding theory, and computer graphics. Factoring polynomials is essential for designing secure encryption algorithms and error-correcting codes. Polynomial interpolation and curve fitting techniques also rely on polynomial factorization.

Economics

In economics, polynomials are used to model cost functions, revenue functions, and profit functions. Factoring these polynomials can help economists analyze the behavior of markets, determine break-even points, and optimize production levels. Understanding the roots of polynomial equations is crucial for making informed business decisions.

Conclusion

Determining which polynomial has $(3x + 2)$ as a binomial factor involves applying the factor theorem and carefully evaluating each option. By substituting the root of the binomial into each polynomial, we can quickly identify whether the result is zero, indicating that the binomial is indeed a factor. In this case, Option D, $21x^4 + 14x^3 + 6x + 4$, is the correct answer.

Mastering polynomial factorization is a crucial skill in algebra and has far-reaching applications in various fields. By understanding the underlying principles and practicing different methods, you can confidently tackle complex polynomial problems and gain a deeper appreciation for the power of algebra.

This comprehensive guide has provided a step-by-step solution to the problem, along with insights into the factor theorem, common pitfalls, alternative methods, and real-world applications of polynomial factorization. Whether you are a student preparing for an exam or a math enthusiast seeking to expand your knowledge, this article serves as a valuable resource for understanding and applying polynomial concepts.