Factorising The Quadratic Expression V^2 + 5v - 6 A Step-by-Step Guide

July 9, 2025 by Scholario Team 71 views

In the realm of mathematics, factorising quadratic expressions is a fundamental skill with applications spanning algebra, calculus, and beyond. This article delves into the process of factorising the quadratic expression $v^2 + 5v - 6$ fully. We will explore the underlying principles, step-by-step methods, and the rationale behind each step. Whether you're a student grappling with algebra or a seasoned mathematician seeking a refresher, this guide aims to provide a comprehensive understanding of factorisation.

Understanding Quadratic Expressions

Before we dive into the specifics of $v^2 + 5v - 6$ , let's establish a solid foundation by defining what quadratic expressions are and why they matter. A quadratic expression is a polynomial of degree two, meaning 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 expression $v^2 + 5v - 6$ fits this form, with $a = 1$ , $b = 5$ , and $c = -6$ . Quadratic expressions are ubiquitous in mathematics and physics, appearing in equations describing projectile motion, areas of geometric shapes, and various other phenomena. Their importance stems from the fact that many real-world problems can be modeled using quadratic equations, making the ability to manipulate and solve them crucial.

The Significance of Factorisation

Factorisation is the process of breaking down a quadratic expression into a product of two linear expressions. In other words, we aim to rewrite $v^2 + 5v - 6$ in the form $(v + p)(v + q)$ , where 'p' and 'q' are constants. Why is this important? Factorisation provides a powerful tool for solving quadratic equations. When a quadratic expression is factorised and set equal to zero, we can easily find the values of the variable that make the equation true. These values, known as the roots or solutions of the equation, have significant practical interpretations in various contexts. For instance, in physics, the roots of a quadratic equation might represent the time at which a projectile hits the ground. Furthermore, factorisation simplifies complex algebraic expressions, making them easier to work with in subsequent calculations. It also aids in understanding the behavior of quadratic functions, such as their intercepts and turning points.

Methods of Factorisation

Several methods exist for factorising quadratic expressions, each with its own advantages and applicability. The most common techniques include:

Trial and Error: This method involves systematically testing different combinations of factors until the correct pair is found. While it can be effective for simple expressions, it becomes less efficient as the complexity increases.
The AC Method: This method, also known as factorisation by grouping, provides a more structured approach. It involves finding two numbers that multiply to 'ac' and add up to 'b'. These numbers are then used to split the middle term, allowing for factorisation by grouping.
Using the Quadratic Formula: This formula provides a direct way to find the roots of a quadratic equation, which can then be used to construct the factors. While it's a reliable method, it can be more time-consuming than other techniques.

In this article, we will focus on the AC method as it offers a systematic and efficient way to factorise $v^2 + 5v - 6$ .

Now, let's apply the AC method to factorise the expression $v^2 + 5v - 6$ step by step.

Step 1: Identify a, b, and c

The first step is to identify the coefficients 'a', 'b', and 'c' in the quadratic expression. As mentioned earlier, in $v^2 + 5v - 6$ , we have:

$a = 1$ (the coefficient of $v^2$ )
$b = 5$ (the coefficient of $v$ )
$c = -6$ (the constant term)

These values are crucial for the subsequent steps in the AC method. Ensuring you correctly identify these coefficients is paramount, as any error here will propagate through the rest of the factorisation process.

Step 2: Calculate ac

The next step is to calculate the product of 'a' and 'c', which is denoted as 'ac'. In our case, $ac = (1) imes (-6) = -6$ . This value is the key to finding the two numbers that will allow us to split the middle term. The product 'ac' represents the target product that the two numbers must multiply to, and it sets the stage for the crucial step of identifying the correct pair of factors.

Step 3: Find Two Numbers That Multiply to ac and Add to b

This is the heart of the AC method. We need to find two numbers that satisfy two conditions:

Their product must be equal to 'ac' (which is -6 in our case).
Their sum must be equal to 'b' (which is 5 in our case).

This step often involves some trial and error, but a systematic approach can make the process more efficient. Start by listing pairs of factors of 'ac' and checking if their sum equals 'b'. For -6, the possible pairs of factors are:

1 and -6 (sum is -5)
-1 and 6 (sum is 5)
2 and -3 (sum is -1)
-2 and 3 (sum is 1)

From this list, we can see that the pair -1 and 6 satisfies both conditions: $(-1) imes (6) = -6$ and $(-1) + (6) = 5$ . These two numbers are the key to splitting the middle term in the next step. Identifying the correct pair of numbers is crucial, as it ensures that the subsequent grouping process will lead to a successful factorisation.

Step 4: Split the Middle Term

Now that we have found the two numbers (-1 and 6), we can split the middle term (5v) in the original expression. We rewrite $v^2 + 5v - 6$ as $v^2 - 1v + 6v - 6$ . Notice that we have replaced 5v with -1v + 6v, which is equivalent. This step is crucial because it sets up the expression for factorisation by grouping, which is the next step in the process. Splitting the middle term correctly ensures that the resulting terms can be grouped in a way that reveals common factors.

Step 5: Factor by Grouping

With the middle term split, we can now factor by grouping. We group the first two terms and the last two terms: $(v^2 - 1v) + (6v - 6)$ . Next, we factor out the greatest common factor (GCF) from each group.

From the first group ( $v^2 - 1v$ ), the GCF is 'v'. Factoring out 'v' gives us $v(v - 1)$ .
From the second group ( $6v - 6$ ), the GCF is 6. Factoring out 6 gives us $6(v - 1)$ .

Now our expression looks like this: $v(v - 1) + 6(v - 1)$ . Notice that both terms have a common factor of $(v - 1)$ . This is a critical observation, as it allows us to factor out $(v - 1)$ from the entire expression.

Step 6: Factor out the Common Factor

Finally, we factor out the common factor $(v - 1)$ from the expression $v(v - 1) + 6(v - 1)$ . This gives us $(v - 1)(v + 6)$ . We have now successfully factorised the quadratic expression $v^2 + 5v - 6$ into the product of two linear expressions. This step represents the culmination of the factorisation process, where the original quadratic expression is rewritten in its factored form.

Therefore, the fully factorised form of $v^2 + 5v - 6$ is $(v - 1)(v + 6)$ .

To ensure our factorisation is correct, we can expand the factored expression and check if it matches the original quadratic expression. Expanding $(v - 1)(v + 6)$ using the distributive property (or the FOIL method) gives us:

$(v - 1)(v + 6) = v(v + 6) - 1(v + 6) = v^2 + 6v - v - 6 = v^2 + 5v - 6$

Since the expanded form matches the original expression, we can confidently conclude that our factorisation is correct. Verification is a crucial step in the factorisation process, as it provides a check against potential errors and ensures the accuracy of the result.

In this article, we have successfully factorised the quadratic expression $v^2 + 5v - 6$ fully using the AC method. We have explored the significance of factorisation, the steps involved in the AC method, and the importance of verification. By mastering these concepts, you will be well-equipped to tackle a wide range of factorisation problems in mathematics and beyond. Factorisation is a cornerstone of algebraic manipulation, and a strong understanding of this skill will undoubtedly benefit you in your mathematical journey. The ability to factorise quadratic expressions not only simplifies algebraic problems but also enhances your problem-solving skills in various mathematical contexts.