Is (x-5) A Factor Of 2x^2 - 7x - 15?

Determining whether a given expression is a factor of a polynomial is a fundamental concept in algebra. In this article, we will delve into the relationship between $(x-5)$ and the polynomial $2x^2 - 7x - 15$. We'll explore various methods to ascertain if $(x-5)$ is indeed a factor, providing a comprehensive understanding of the underlying principles. Our main focus keywords are factor and polynomial, crucial terms in this exploration.

Understanding Factors and Polynomials

Before we dive into the specifics of this problem, let's clarify the terms we're working with. A polynomial is an expression consisting of variables and coefficients, combined using addition, subtraction, and multiplication, with non-negative integer exponents. Examples include $x^2 + 3x - 2$, $5x^4 - 7x^2 + 1$, and, of course, our polynomial of interest, $2x^2 - 7x - 15$. A factor, on the other hand, is an expression that divides evenly into another expression. In simpler terms, if we can divide a polynomial by a certain expression and get a remainder of zero, then that expression is a factor of the polynomial. For example, $(x+2)$ is a factor of $x^2 + 5x + 6$ because $(x^2 + 5x + 6) / (x+2) = x + 3$ with no remainder. Understanding this relationship between factors and polynomials is the cornerstone of our investigation.

Methods to Determine Factors

There are several methods we can employ to determine if $(x-5)$ is a factor of $2x^2 - 7x - 15$. We will primarily focus on two methods: the factor theorem and polynomial long division. Each method offers a unique approach to the problem, and understanding both will provide a more robust understanding of factor identification. The factor theorem offers a quick way to check if a linear expression is a factor, while polynomial long division provides a more comprehensive view, even yielding the other factor if it exists. By mastering these methods, you'll be well-equipped to tackle similar problems involving polynomials and their factors.

Method 1: The Factor Theorem

The Factor Theorem is a powerful tool that provides a direct way to check if a linear expression $(x - a)$ is a factor of a polynomial $P(x)$. The theorem states that $(x - a)$ is a factor of $P(x)$ if and only if $P(a) = 0$. In simpler terms, we substitute the value that makes the factor equal to zero into the polynomial. If the result is zero, then the expression is indeed a factor. This method offers a shortcut, saving us from the more laborious process of long division in certain cases. This theorem is deeply rooted in the relationship between roots and factors of polynomials. Finding a root of a polynomial is equivalent to finding a factor, and the Factor Theorem formalizes this connection.

Applying the Factor Theorem

In our case, we want to check if $(x - 5)$ is a factor of $2x^2 - 7x - 15$. According to the Factor Theorem, we need to find the value that makes $(x - 5)$ equal to zero. Solving the equation $x - 5 = 0$, we find that $x = 5$. Now, we substitute $x = 5$ into the polynomial $P(x) = 2x^2 - 7x - 15$:

$P(5) = 2(5)^2 - 7(5) - 15$

$P(5) = 2(25) - 35 - 15$

$P(5) = 50 - 35 - 15$

$P(5) = 0$

Since $P(5) = 0$, the Factor Theorem tells us that $(x - 5)$ is indeed a factor of $2x^2 - 7x - 15$. This elegant method provides a definitive answer with minimal computation, highlighting the power of the Factor Theorem in factor analysis. The fact that the polynomial evaluates to zero when $x = 5$ strongly suggests a direct relationship between the root and the corresponding linear factor.

Method 2: Polynomial Long Division

Polynomial long division is another method to determine if $(x - 5)$ is a factor of $2x^2 - 7x - 15$. This method is similar to long division with numbers, but instead of dividing numbers, we divide polynomials. If the remainder after the division is zero, then the divisor is a factor of the dividend. Polynomial long division is a fundamental skill in algebra, useful not only for determining factors but also for simplifying rational expressions and solving polynomial equations. It provides a visual and systematic way to perform division, making it easier to track each step and identify any potential errors.

Performing Polynomial Long Division

To perform polynomial long division, we set up the division problem as follows:

 x - 5 | 2x^2 - 7x - 15

First, we divide the leading term of the dividend ($2x^2$) by the leading term of the divisor ($x$), which gives us $2x$. We write $2x$ above the $-7x$ term.

 2x
 x - 5 | 2x^2 - 7x - 15

Next, we multiply the divisor $(x - 5)$ by $2x$, which gives us $2x^2 - 10x$. We write this below the dividend and subtract:

 2x
 x - 5 | 2x^2 - 7x - 15
 -(2x^2 - 10x)

This simplifies to:

 2x
 x - 5 | 2x^2 - 7x - 15
 - (2x^2 - 10x)
 3x - 15

Now, we bring down the next term from the dividend, which is $-15$. We then divide the leading term of the new expression ($3x$) by the leading term of the divisor ($x$), which gives us $3$. We write $+3$ next to the $2x$ above.

 2x + 3
 x - 5 | 2x^2 - 7x - 15
 - (2x^2 - 10x)
 3x - 15

We multiply the divisor $(x - 5)$ by $3$, which gives us $3x - 15$. We write this below the $3x - 15$ and subtract:

 2x + 3
 x - 5 | 2x^2 - 7x - 15
 - (2x^2 - 10x)
 3x - 15
 - (3x - 15)

This simplifies to:

 2x + 3
 x - 5 | 2x^2 - 7x - 15
 - (2x^2 - 10x)
 3x - 15
 - (3x - 15)
 0

Since the remainder is $0$, we conclude that $(x - 5)$ is a factor of $2x^2 - 7x - 15$. Moreover, the quotient $2x + 3$ is the other factor. This process demonstrates the effectiveness of polynomial long division in not only verifying factors but also in finding the complementary factor. The zero remainder is a clear indication of complete divisibility, confirming the factor relationship.

Conclusion: (x-5) is a Factor

Both the Factor Theorem and polynomial long division confirm that $(x - 5)$ is indeed a factor of the polynomial $2x^2 - 7x - 15$. The Factor Theorem provided a quick and elegant verification by showing that $P(5) = 0$. Polynomial long division offered a more detailed approach, not only confirming that $(x - 5)$ is a factor but also revealing the other factor, which is $(2x + 3)$. Understanding these methods and the relationship between factors and polynomials is crucial for success in algebra. By applying these techniques, you can confidently determine whether a given expression is a factor of a polynomial and further explore the properties of polynomials. This exploration highlights the interconnectedness of algebraic concepts and the power of different techniques in solving the same problem.

Therefore, the correct answer is B. $(x-5)$ is a factor.