Factoring $y^2 + \frac{b^2-1}{b}y - 1$ A Step-by-Step Guide

Introduction

In this article, we will delve into the process of factoring the quadratic expression $y^2 + \frac{b^2-1}{b}y - 1$. This type of problem often appears in algebra and calculus, and mastering the technique of factoring is crucial for solving equations and simplifying expressions. We will break down the steps involved, providing a comprehensive guide for anyone looking to understand and solve this kind of problem. Our primary goal is to find two binomials that, when multiplied together, yield the original quadratic expression. This involves identifying the correct factors and ensuring they align with the coefficients in the expression. Factoring not only helps in solving quadratic equations but also provides a deeper understanding of the underlying algebraic structure. So, let’s embark on this journey and demystify the process of factoring this specific quadratic expression.

Understanding Quadratic Expressions

Before we dive into the specifics of factoring $y^2 + \frac{b^2-1}{b}y - 1$, it's essential to have a solid grasp of quadratic expressions in general. A quadratic expression is a polynomial of degree two, which means the highest power of the variable is two. The general form of a quadratic expression is $ax^2 + bx + c$, where $a$, $b$, and $c$ are constants, and $x$ is the variable. In our case, the variable is $y$, and the expression is $y^2 + \frac{b^2-1}{b}y - 1$. Here, $a = 1$, $b = \frac{b^2-1}{b}$, and $c = -1$. Understanding these coefficients is crucial for factoring the expression correctly.

The process of factoring a quadratic expression involves breaking it down into a product of two binomials. A binomial is a polynomial with two terms, such as $(y + m)$ and $(y + n)$. When we multiply these binomials, we should obtain the original quadratic expression. This means we are looking for two numbers, let's call them $m$ and $n$, such that $(y + m)(y + n) = y^2 + \frac{b^2-1}{b}y - 1$. The key to successful factoring lies in finding the correct values for $m$ and $n$. These values must satisfy two conditions: their product ($m imes n$) must equal the constant term $c$, and their sum ($m + n$) must equal the coefficient of the linear term $b$. In our example, $m imes n$ should equal $-1$, and $m + n$ should equal $\frac{b^2-1}{b}$. This might seem challenging at first, but with practice and a systematic approach, it becomes much more manageable. Factoring is a fundamental skill in algebra and serves as a building block for more advanced mathematical concepts.

Setting Up the Problem

Now, let's set up the specific problem we aim to solve: factoring the quadratic expression $y^2 + \frac{b^2-1}{b}y - 1$. As we discussed earlier, our goal is to find two binomials of the form $(y + m)(y + n)$ that multiply together to give us the original quadratic expression. This means we need to find the values of $m$ and $n$ that satisfy the conditions $m imes n = -1$ and $m + n = \frac{b^2-1}{b}$. The first condition, $m imes n = -1$, tells us that one of the numbers, $m$ or $n$, must be positive, and the other must be negative. This is because the product of two numbers is negative only if they have opposite signs. The second condition, $m + n = \frac{b^2-1}{b}$, gives us a relationship between the two numbers and the variable $b$. This is where the problem becomes more interesting and requires careful consideration.

To tackle this, let's rewrite the second condition in a more useful form. We can split the fraction $\frac{b^2-1}{b}$ into two separate fractions: $\frac{b^2}{b} - \frac{1}{b}$, which simplifies to $b - \frac{1}{b}$. So, we now have $m + n = b - \frac{1}{b}$. This tells us that the sum of our two numbers should be equal to $b$ minus the reciprocal of $b$. With these two conditions in mind—$m imes n = -1$ and $m + n = b - \frac{1}{b}$—we can start exploring possible values for $m$ and $n$. The key is to find two numbers that not only multiply to $-1$ but also add up to $b - \frac{1}{b}$. This might involve some trial and error, but a systematic approach will help us narrow down the possibilities and arrive at the correct factors.

Finding the Factors

To find the factors for the quadratic expression $y^2 + \frac{b^2-1}{b}y - 1$, we need to identify two numbers, $m$ and $n$, that satisfy the two crucial conditions we established earlier: $m imes n = -1$ and $m + n = b - \frac{1}{b}$. The first condition, $m imes n = -1$, significantly narrows down our options. Since the product of $m$ and $n$ is $-1$, we know that one number must be $1$ and the other must be $-1$, but with appropriate signs and potentially scaled by factors involving $b$. This is because $1$ and $-1$ are the only integers that multiply to $-1$. However, we need to consider that $m$ and $n$ could also be fractions or expressions involving $b$.

Given the second condition, $m + n = b - \frac{1}{b}$, we can start to piece together the possible values of $m$ and $n$. Let's consider the terms $b$ and $-\frac{1}{b}$ separately. If we let $m = b$ and $n = -\frac{1}{b}$, we can see if these values satisfy both conditions. First, let's check the product: $m imes n = b imes (-\frac{1}{b}) = -1$. This satisfies our first condition. Now, let's check the sum: $m + n = b + (-\frac{1}{b}) = b - \frac{1}{b}$. This also satisfies our second condition. Therefore, we have found the two numbers we were looking for: $m = b$ and $n = -\frac{1}{b}$. These values are the key to factoring the quadratic expression.

Utilizing the Conditions

The conditions we've established, $m imes n = -1$ and $m + n = b - \frac{1}{b}$, are pivotal in finding the correct factors. The condition $m imes n = -1$ is particularly insightful because it tells us that the factors must have opposite signs. This is a direct consequence of the fact that the product of two numbers is negative only if one is positive and the other is negative. This narrows our search considerably, as we don't need to consider pairs of numbers with the same sign. The condition also suggests that the absolute values of $m$ and $n$ are reciprocals of each other, in a sense, since their product is $-1$.

The second condition, $m + n = b - \frac{1}{b}$, provides a more nuanced relationship between the factors and the coefficient of the $y$ term in the quadratic expression. It tells us that the sum of the two factors must equal $b$ minus its reciprocal. This is a crucial piece of information that helps us refine our search for the correct factors. By analyzing this condition, we can make educated guesses about the possible values of $m$ and $n$. For instance, if $b$ is a large positive number, then $b - \frac{1}{b}$ will also be a large positive number, suggesting that the positive factor will be larger in magnitude than the negative factor. Conversely, if $b$ is a small positive number close to zero, then $-\frac{1}{b}$ will be a large negative number, suggesting that the negative factor will be larger in magnitude than the positive factor.

By carefully considering both conditions together, we can systematically narrow down the possibilities and identify the correct factors. It's a bit like solving a puzzle, where each condition provides a piece of the solution. The more we understand these conditions, the better equipped we are to factor quadratic expressions efficiently and accurately. This methodical approach is a valuable skill in algebra and beyond, as it teaches us to break down complex problems into manageable parts and use logical reasoning to arrive at the solution.

Constructing the Factored Form

Now that we have identified the factors $m = b$ and $n = -\frac{1}{b}$, we can proceed to construct the factored form of the quadratic expression $y^2 + \frac{b^2-1}{b}y - 1$. Recall that our goal was to find two binomials of the form $(y + m)(y + n)$ that multiply together to give us the original quadratic expression. Since we have found the values of $m$ and $n$, we can simply substitute them into the binomials.

Substituting $m = b$ and $n = -\frac{1}{b}$, we get the binomials $(y + b)$ and $(y - \frac{1}{b})$. Therefore, the factored form of the quadratic expression is $(y + b)(y - \frac{1}{b})$. To verify that this is indeed the correct factorization, we can multiply the two binomials together and see if we obtain the original quadratic expression. When we multiply $(y + b)(y - \frac{1}{b})$, we use the distributive property (also known as the FOIL method) to expand the product:

$(y + b)(y - \frac{1}{b}) = y(y - \frac{1}{b}) + b(y - \frac{1}{b}) = y^2 - \frac{y}{b} + by - \frac{b}{b} = y^2 + (b - \frac{1}{b})y - 1$

This is exactly the original quadratic expression, $y^2 + \frac{b^2-1}{b}y - 1$, which confirms that our factorization is correct. The factored form $(y + b)(y - \frac{1}{b})$ provides a concise and useful representation of the quadratic expression. It allows us to easily identify the roots of the corresponding quadratic equation (by setting each factor equal to zero) and provides insights into the behavior of the quadratic function.

Verification Process

The verification process is a crucial step in factoring any quadratic expression, as it ensures that the factored form is equivalent to the original expression. In our case, we factored $y^2 + \frac{b^2-1}{b}y - 1$ into $(y + b)(y - \frac{1}{b})$. To verify this factorization, we need to multiply the two binomials together and check if the result matches the original expression. This process involves applying the distributive property, often remembered by the acronym FOIL (First, Outer, Inner, Last).

Let's break down the multiplication step by step. We start by multiplying the first terms in each binomial: $y imes y = y^2$. Next, we multiply the outer terms: $y imes (-\frac{1}{b}) = -\frac{y}{b}$. Then, we multiply the inner terms: $b imes y = by$. Finally, we multiply the last terms: $b imes (-\frac{1}{b}) = -1$. Now, we add all these products together:

$y^2 - \frac{y}{b} + by - 1$

To simplify this expression and make it easier to compare with the original, we can group the terms involving $y$:

$y^2 + (b - \frac{1}{b})y - 1$

Now, we can see that this is exactly the same as the original quadratic expression, $y^2 + \frac{b^2-1}{b}y - 1$, since $b - \frac{1}{b}$ is equivalent to $\frac{b^2-1}{b}$. This confirms that our factorization is correct. The verification process not only gives us confidence in our solution but also reinforces our understanding of the factoring process. It's a valuable habit to develop, as it helps prevent errors and ensures accuracy in algebraic manipulations.

The Solution

Having gone through the process of finding the factors and constructing the factored form, we have arrived at the solution for factoring the quadratic expression $y^2 + \frac{b^2-1}{b}y - 1$. The solution is the factored form $(y + b)(y - \frac{1}{b})$. This means that the original quadratic expression can be written as the product of these two binomials. We have verified this factorization by multiplying the binomials together and showing that the result is indeed the original expression.

This solution provides a concise and insightful representation of the quadratic expression. It allows us to easily identify the roots of the corresponding quadratic equation, which are the values of $y$ that make the expression equal to zero. To find the roots, we simply set each factor equal to zero and solve for $y$:

• $y + b = 0$ implies $y = -b$
• $y - \frac{1}{b} = 0$ implies $y = \frac{1}{b}$

So, the roots of the quadratic equation $y^2 + \frac{b^2-1}{b}y - 1 = 0$ are $y = -b$ and $y = \frac{1}{b}$. This illustrates one of the key benefits of factoring: it simplifies the process of finding the roots of a quadratic equation. The factored form also gives us insights into the graph of the quadratic function. The roots correspond to the x-intercepts of the graph, and the factored form helps us understand the symmetry and behavior of the parabola.

Roots of the Equation

Furthermore, understanding the roots of the equation provides a deeper understanding of the quadratic expression and its behavior. The roots, as we determined, are $y = -b$ and $y = \frac{1}{b}$. These values are significant because they are the points where the graph of the quadratic function $y^2 + \frac{b^2-1}{b}y - 1$ intersects the x-axis. In other words, they are the solutions to the equation $y^2 + \frac{b^2-1}{b}y - 1 = 0$.

The roots of a quadratic equation have several important properties. For example, their sum is equal to the negative of the coefficient of the $y$ term, which in this case is $-(b - \frac{1}{b}) = \frac{1}{b} - b$. If we add the roots we found, $-b$ and $\frac{1}{b}$, we indeed get $\frac{1}{b} - b$, which confirms this property. Similarly, the product of the roots is equal to the constant term, which is $-1$. Multiplying our roots, $(-b) imes (\frac{1}{b})$, gives us $-1$, further validating our solution.

Knowing the roots of a quadratic equation is also crucial for sketching the graph of the corresponding quadratic function. The roots tell us where the parabola crosses the x-axis, and the sign of the coefficient of the $y^2$ term (which is $1$ in this case, indicating a positive value) tells us whether the parabola opens upwards or downwards. Since the coefficient is positive, the parabola opens upwards. With the roots and the direction of opening, we can get a good sense of the shape and position of the parabola. Understanding the roots is not just about solving an equation; it's about gaining a comprehensive understanding of the underlying mathematical concepts and their graphical representations.

Conclusion

In conclusion, we have successfully factored the quadratic expression $y^2 + \frac{b^2-1}{b}y - 1$ into its factored form, $(y + b)(y - \frac{1}{b})$. This process involved understanding the general form of quadratic expressions, identifying the key conditions for factoring, systematically finding the factors, constructing the factored form, and verifying our solution. We also explored the significance of the roots of the corresponding quadratic equation, which are $y = -b$ and $y = \frac{1}{b}$.

Factoring quadratic expressions is a fundamental skill in algebra, with applications in various areas of mathematics and science. It not only helps in solving equations but also provides valuable insights into the structure and behavior of quadratic functions. The systematic approach we followed in this article can be applied to factor other quadratic expressions as well. The key is to carefully analyze the coefficients, identify the conditions that the factors must satisfy, and use a combination of trial and error and logical reasoning to find the correct factors.

This article has provided a comprehensive guide to factoring a specific quadratic expression, but the principles and techniques discussed are applicable to a wide range of similar problems. By mastering these skills, you will be well-equipped to tackle more advanced algebraic concepts and solve complex mathematical problems. The journey of understanding mathematics is like building a strong foundation, where each concept learned serves as a building block for the next. Factoring quadratic expressions is one such crucial building block, and we hope this article has helped you strengthen your understanding of this important topic.