Multiplying Polynomials Using FOIL Method - Step-by-Step Guide

Multiplying binomials can seem daunting at first, but the FOIL method provides a systematic approach to ensure accuracy. FOIL stands for First, Outer, Inner, Last, representing the order in which terms from the two binomials are multiplied. In this article, we will delve into the FOIL method and apply it to the expression $5(y-6)(3y-1)$. This comprehensive guide will not only walk you through the step-by-step process but also ensure you understand the underlying principles of binomial multiplication. By mastering this method, you'll be well-equipped to tackle more complex algebraic expressions and confidently solve a variety of mathematical problems. Understanding the nuances of the FOIL method is crucial for students and anyone dealing with algebraic manipulations. It’s a foundational skill that simplifies the multiplication of binomials, making it easier to expand and simplify expressions. The method itself is an acronym that serves as a mnemonic device to remember the correct order of operations: First, Outer, Inner, and Last. Each letter corresponds to a specific pair of terms you need to multiply. First, you multiply the first terms in each binomial. Outer, you multiply the outermost terms in the expression. Inner, you multiply the innermost terms. Last, you multiply the last terms in each binomial. Once you've performed these multiplications, you combine like terms to simplify the resulting expression. This systematic approach ensures that you don't miss any terms and that the multiplication is carried out correctly. In the sections that follow, we will break down the application of the FOIL method to the expression $5(y-6)(3y-1)$, providing a detailed, step-by-step explanation that will make the process clear and understandable.

Understanding the FOIL Method

To effectively use the FOIL method, let's first break down what each letter represents. F stands for First, meaning you multiply the first terms of each binomial. O stands for Outer, indicating you multiply the outermost terms. I stands for Inner, meaning you multiply the innermost terms, and L stands for Last, where you multiply the last terms of each binomial. This structured approach ensures that you account for every possible term combination when multiplying two binomials. The FOIL method is a straightforward technique designed to simplify the multiplication of binomials. By following the acronym, you ensure that every term in the first binomial is correctly multiplied by every term in the second binomial. This is crucial for arriving at the correct expanded form of the expression. The First step involves multiplying the first term of the first binomial by the first term of the second binomial. This is often the most straightforward part of the process, but it's important to get it right. The Outer step focuses on the terms that are farthest apart in the expression – the first term of the first binomial and the last term of the second binomial. This multiplication adds another layer to the expanded expression. The Inner step deals with the two terms that are closest together – the second term of the first binomial and the first term of the second binomial. This step is just as crucial as the others in ensuring the accuracy of the final result. Finally, the Last step involves multiplying the last terms of each binomial. Once you've completed all four steps, you'll have a series of terms that need to be combined and simplified. The key to mastering the FOIL method is consistent practice. By working through various examples, you'll become more comfortable with the process and more adept at identifying and combining like terms. This will not only improve your algebraic skills but also provide a solid foundation for more advanced mathematical concepts. In the subsequent sections, we'll apply this method to the specific expression $5(y-6)(3y-1)$, demonstrating how each step is executed in practice.

Applying FOIL to $(y-6)(3y-1)$

Let's apply the FOIL method to the binomials $(y-6)$ and $(3y-1)$. First, we multiply the First terms: $y * 3y = 3y^2$. Next, we multiply the Outer terms: $y * -1 = -y$. Then, we multiply the Inner terms: $-6 * 3y = -18y$. Finally, we multiply the Last terms: $-6 * -1 = 6$. This gives us the expanded form: $3y^2 - y - 18y + 6$. Before we can incorporate the factor of 5, it's crucial to properly expand the binomials $(y-6)$ and $(3y-1)$. This involves meticulously applying the FOIL method to ensure no term is missed and that the multiplications are performed accurately. The FOIL method, as we discussed, stands for First, Outer, Inner, Last, and it provides a structured approach to multiplying two binomials. Starting with the First terms, we multiply $y$ (from the first binomial) by $3y$ (from the second binomial). This results in $3y^2$. This is a fundamental step, setting the stage for the rest of the expansion. Moving on to the Outer terms, we multiply $y$ (from the first binomial) by $-1$ (from the second binomial), resulting in $-y$. This step is equally important, as it accounts for the interaction between the outer terms of the two binomials. Next, we focus on the Inner terms. Here, we multiply $-6$ (from the first binomial) by $3y$ (from the second binomial), which gives us $-18y$. This multiplication captures the relationship between the inner terms and is essential for a complete expansion. Finally, we address the Last terms. We multiply $-6$ (from the first binomial) by $-1$ (from the second binomial), resulting in $6$. This completes the multiplication process according to the FOIL method. Combining these results, we get the expanded form $3y^2 - y - 18y + 6$. This expression is not yet fully simplified, as we still need to combine like terms. However, the FOIL method has successfully expanded the product of the two binomials, laying the groundwork for the final simplification. In the next step, we'll combine the like terms and then consider the effect of the factor of 5.

Combining Like Terms

Now, let's combine like terms in the expression $3y^2 - y - 18y + 6$. We can combine the '-y' and '-18y' terms, which gives us $-19y$. So, the expression becomes $3y^2 - 19y + 6$. Combining like terms is a critical step in simplifying algebraic expressions, and it’s particularly important after applying the FOIL method. The expression we've obtained, $3y^2 - y - 18y + 6$, has two terms that can be combined: $-y$ and $-18y$. These are like terms because they both contain the variable $y$ raised to the same power (in this case, $y$ to the power of 1). To combine them, we simply add their coefficients. The coefficient of $-y$ is $-1$, and the coefficient of $-18y$ is $-18$. Adding these together, we get $-1 + (-18) = -19$. Therefore, the combined term is $-19y$. Now, we replace the original terms $-y$ and $-18y$ with the combined term $-19y$ in the expression. This gives us the simplified expression $3y^2 - 19y + 6$. This expression is now in a more compact and manageable form. It consists of three terms: a quadratic term ($3y^2$), a linear term ($-19y$), and a constant term ($6$). None of these terms can be further combined because they are not like terms. The quadratic term has $y$ raised to the power of 2, the linear term has $y$ raised to the power of 1, and the constant term has no $y$ at all. At this point, we've successfully simplified the expression that resulted from applying the FOIL method to the binomials $(y-6)$ and $(3y-1)$. However, we still need to account for the factor of 5 that was present in the original expression: $5(y-6)(3y-1)$. In the next step, we'll distribute this factor of 5 across the simplified trinomial $3y^2 - 19y + 6$ to complete the problem.

Distributing the 5

Finally, we need to distribute the 5 across the expression $3y^2 - 19y + 6$. This means multiplying each term in the expression by 5: $5 * 3y^2 = 15y^2$, $5 * -19y = -95y$, and $5 * 6 = 30$. So, the final expression is $15y^2 - 95y + 30$. Distributing the 5 is the final step in simplifying the expression $5(y-6)(3y-1)$. This step involves multiplying each term within the trinomial $3y^2 - 19y + 6$ by the factor of 5. This ensures that every term in the expanded expression is properly scaled by the initial factor. Starting with the first term, we multiply $5$ by $3y^2$. This gives us $5 * 3y^2 = 15y^2$. This is a straightforward multiplication, where we simply multiply the coefficients. Next, we multiply $5$ by the second term, $-19y$. This gives us $5 * -19y = -95y$. Again, we multiply the coefficients, paying attention to the negative sign. Finally, we multiply $5$ by the constant term, $6$. This gives us $5 * 6 = 30$. This completes the distribution process. Now, we combine the results of these multiplications to form the final simplified expression. We have $15y^2$, $-95y$, and $30$. Putting these together, we get $15y^2 - 95y + 30$. This is the fully expanded and simplified form of the original expression $5(y-6)(3y-1)$. It's a quadratic expression in the standard form $ax^2 + bx + c$, where $a = 15$, $b = -95$, and $c = 30$. By systematically applying the FOIL method, combining like terms, and distributing the constant factor, we've successfully transformed the initial expression into its simplified form. This demonstrates the power and utility of these algebraic techniques in manipulating and simplifying expressions. The final answer, $15y^2 - 95y + 30$, represents the solution to the problem and highlights the importance of each step in the process. In conclusion, multiplying binomials and simplifying expressions involves a series of steps, each requiring careful attention to detail. From applying the FOIL method to combining like terms and distributing constants, a systematic approach is key to arriving at the correct answer. By mastering these techniques, you can confidently tackle a wide range of algebraic problems and build a solid foundation for more advanced mathematical concepts.

Final Answer

Therefore, $5(y-6)(3y-1) = 15y^2 - 95y + 30$.