Is (y-1) A Factor Of Y³ - 9y² + 10y + 5? Explained

Hey there, math enthusiasts! Ever wondered how to determine if a binomial like (y-1) is a factor of a polynomial? It might seem daunting at first, but with the right approach, it's totally manageable. In this article, we'll dive deep into this question: Is (y-1) a factor of the polynomial y³ - 9y² + 10y + 5? We'll explore different methods, break down the concepts, and make sure you walk away with a solid understanding. So, let's get started!

Understanding Factors and Polynomials

Before we jump into the problem, let’s get our basics straight. Understanding factors and polynomials is crucial. A polynomial is essentially an expression consisting of variables and coefficients, combined using addition, subtraction, and multiplication, with non-negative integer exponents. Think of it as a mathematical Swiss Army knife – versatile and used in many areas of math and science. For example, y³ - 9y² + 10y + 5 is a polynomial. The degree of this polynomial is 3 because the highest power of the variable 'y' is 3. The coefficients are the numbers that multiply the variables (or the constant term, which doesn't have a variable). In our example, the coefficients are 1 (for y³), -9 (for y²), 10 (for y), and 5 (the constant term).

Now, what about factors? A factor is a number or expression that divides another number or expression evenly, with no remainder. For instance, 2 is a factor of 6 because 6 divided by 2 equals 3, with no remainder. Similarly, in the world of polynomials, a factor is another polynomial that divides the given polynomial evenly. If we say (y - 1) is a factor of y³ - 9y² + 10y + 5, it means that when we divide y³ - 9y² + 10y + 5 by (y - 1), we should get another polynomial with no remainder. This concept is pivotal in simplifying complex expressions and solving equations, making it a cornerstone of algebra and beyond.

Method 1: The Factor Theorem

One of the most elegant ways to determine if (y-1) is a factor is by using the Factor Theorem. The Factor Theorem is a powerful tool that links factors of a polynomial to its roots. In simple terms, it states that if a polynomial f(y) has a factor (y - a), then f(a) must equal zero. Conversely, if f(a) equals zero, then (y - a) is a factor of f(y). It’s like a mathematical detective, helping us uncover hidden factors!

So, how do we apply this to our problem? Our polynomial is f(y) = y³ - 9y² + 10y + 5, and we want to check if (y - 1) is a factor. According to the Factor Theorem, we need to evaluate f(1). This means we substitute y = 1 into our polynomial: f(1) = (1)³ - 9(1)² + 10(1) + 5. Let's break it down: (1)³ is 1, 9(1)² is 9, 10(1) is 10. So, f(1) = 1 - 9 + 10 + 5. Now, let's add and subtract: 1 - 9 = -8, -8 + 10 = 2, and 2 + 5 = 7. Therefore, f(1) = 7.

Since f(1) equals 7, which is not zero, we can conclude that (y - 1) is not a factor of y³ - 9y² + 10y + 5. The Factor Theorem has given us a straightforward answer without having to perform long division or other complex calculations. This method highlights the theorem's efficiency in quickly determining factors of polynomials.

Method 2: Polynomial Long Division

Another classic method to figure this out is Polynomial Long Division. Think of it like the long division you learned in elementary school, but now we're dealing with polynomials instead of numbers. It's a bit more involved than the Factor Theorem, but it's a solid method for understanding how polynomials divide and whether there's a remainder.

Let’s walk through it step by step. We want to divide y³ - 9y² + 10y + 5 by (y - 1). First, set up the long division just like you would with numbers: write the divisor (y - 1) outside the division bracket and the dividend (y³ - 9y² + 10y + 5) inside. Now, focus on the highest degree terms. We ask: What do we need to multiply (y - 1) by to get y³? The answer is y². So, write y² above the division bracket, aligned with the y² term in the dividend. Next, multiply (y - 1) by y² to get y³ - y². Write this below the dividend and subtract: (y³ - 9y² + 10y + 5) - (y³ - y²) = -8y² + 10y + 5.

Bring down the next term (+10y) to get -8y² + 10y. Now, repeat the process. What do we need to multiply (y - 1) by to get -8y²? The answer is -8y. Write -8y above the division bracket, aligned with the y term. Multiply (y - 1) by -8y to get -8y² + 8y. Write this below and subtract: (-8y² + 10y + 5) - (-8y² + 8y) = 2y + 5. Bring down the last term (+5) to get 2y + 5. One last time, what do we need to multiply (y - 1) by to get 2y? The answer is 2. Write +2 above the division bracket, aligned with the constant term. Multiply (y - 1) by 2 to get 2y - 2. Write this below and subtract: (2y + 5) - (2y - 2) = 7.

We are left with a remainder of 7. Since the remainder is not zero, this tells us that (y - 1) is not a factor of y³ - 9y² + 10y + 5. Polynomial Long Division not only gives us the quotient but also definitively shows whether the divisor is a factor by the presence or absence of a remainder.

Method 3: Synthetic Division

For those who love a streamlined approach, Synthetic Division is the way to go. Think of it as a shortcut version of polynomial long division – it's quicker and more efficient, especially for linear divisors like (y - 1). However, it's important to note that synthetic division works best when dividing by a linear factor of the form (y - a). It’s a neat trick to have up your sleeve for specific situations.

Here’s how it works for our problem. First, identify the root of the divisor (y - 1). Setting (y - 1) equal to zero gives us y = 1. This is the number we'll use in our synthetic division setup. Write down the coefficients of the polynomial y³ - 9y² + 10y + 5. These are 1 (for y³), -9 (for y²), 10 (for y), and 5 (the constant term). Set up the synthetic division table: write the root (1) to the left, and the coefficients (1, -9, 10, 5) to the right.

Now, bring down the first coefficient (1) to the bottom row. Multiply this by the root (1) to get 1, and write it under the next coefficient (-9). Add -9 and 1 to get -8, and write this in the bottom row. Multiply -8 by the root (1) to get -8, and write it under the next coefficient (10). Add 10 and -8 to get 2, and write this in the bottom row. Finally, multiply 2 by the root (1) to get 2, and write it under the last coefficient (5). Add 5 and 2 to get 7, which is the remainder.

The last number in the bottom row, 7, is the remainder. The other numbers (1, -8, 2) are the coefficients of the quotient polynomial. Since the remainder is 7, which is not zero, we can definitively say that (y - 1) is not a factor of y³ - 9y² + 10y + 5. Synthetic division provides a quick and clean way to check for factors, particularly useful for linear divisors. It's a powerful tool in polynomial arithmetic.

Conclusion: (y-1) is Not a Factor

So, we've explored three different methods – the Factor Theorem, Polynomial Long Division, and Synthetic Division – and all of them lead us to the same conclusion: (y - 1) is not a factor of y³ - 9y² + 10y + 5. Whether we substituted y = 1 into the polynomial using the Factor Theorem, performed the step-by-step division, or used the streamlined synthetic division, the result was consistent. This reinforces the power and reliability of these algebraic techniques.

Understanding how to determine factors of polynomials is a fundamental skill in algebra and calculus. It allows you to simplify expressions, solve equations, and analyze functions more effectively. The Factor Theorem provides a quick check, Polynomial Long Division offers a comprehensive view of the division process, and Synthetic Division offers a speedy alternative for linear divisors. By mastering these methods, you'll be well-equipped to tackle a wide range of polynomial problems.

Keep practicing, keep exploring, and you'll find that these seemingly complex concepts become second nature. Happy math-ing, guys!