Expressing Polynomials As Products Of Linear Factors A Step-by-Step Guide

Hey guys! Let's dive into how to express the polynomial $2x^4 - 7x^3 - 27x^2 + 63x + 81$ as a product of linear factors. This is a super important skill in algebra, and we're going to break it down step by step so it's crystal clear. We aim to rewrite the given polynomial in the form $(x + ?)(x - 3)(x + 1)(2x - [])$, which involves finding the missing constants. So, let’s get started and make some math magic happen!

Understanding Polynomial Factorization

Before we jump into the nitty-gritty, let's quickly recap what polynomial factorization actually means. When we talk about factoring a polynomial, we're essentially trying to rewrite it as a product of simpler polynomials, ideally linear factors (polynomials of degree one). Think of it like breaking down a number into its prime factors – except we're doing it with polynomials. Why do we do this? Well, factoring helps us find the roots (or zeros) of the polynomial, which are the values of $x$ that make the polynomial equal to zero. These roots are incredibly useful in various applications, from solving equations to graphing functions.

Now, in our specific case, we have the polynomial $2x^4 - 7x^3 - 27x^2 + 63x + 81$, which is a quartic polynomial (degree four). This means it can have up to four roots. Our goal is to express this polynomial as a product of four linear factors. The given form $(x + ?)(x - 3)(x + 1)(2x - [])$ provides a roadmap, indicating that we already know two factors: $(x - 3)$ and $(x + 1)$. This is a great head start! We just need to figure out the missing pieces.

To find the missing constants, we'll often use techniques such as the Rational Root Theorem, synthetic division, and good old-fashioned polynomial division. The Rational Root Theorem helps us identify potential rational roots (roots that can be expressed as fractions), which we can then test using synthetic division. Synthetic division is a streamlined way to divide a polynomial by a linear factor, and it's super handy for checking if a potential root actually works. If the remainder is zero, then we've found a factor! Polynomial division is a more general technique that can be used to divide by any polynomial, not just linear factors, and it will be useful as we progress.

Initial Observations and Strategy

Okay, let’s take a closer look at the polynomial $2x^4 - 7x^3 - 27x^2 + 63x + 81$ and the target factored form $(x + ?)(x - 3)(x + 1)(2x - [])$. We already know that $(x - 3)$ and $(x + 1)$ are factors. This means that $x = 3$ and $x = -1$ are roots of the polynomial. That's valuable information! We can use these roots to simplify the polynomial. Think of each factor as a piece of the puzzle, and our job is to fit all the pieces together.

Notice that the leading coefficient of our polynomial is 2, which comes from the $2x$ term in the last factor $(2x - [])$. This tells us that one of the remaining factors must have a leading coefficient of 2 or a fractional root if we were to find one. This is a crucial observation because it narrows down our possibilities. The constant term of the polynomial is 81, which is the product of all the constant terms in the linear factors. So, the constants we're looking for must be factors of 81. This gives us a finite set of possibilities to test, which is fantastic!

Our strategy will be to use the known roots to reduce the polynomial's degree, making it easier to handle. We'll use synthetic division with $x = 3$ and $x = -1$ to divide the polynomial and see what's left. The result will be a quadratic polynomial, which we can then factor using standard techniques like factoring by grouping, the quadratic formula, or simply recognizing common factors. By methodically working through these steps, we'll uncover the missing constants and write the polynomial as a product of linear factors. Remember, math is like detective work – each clue brings us closer to the solution!

Applying Synthetic Division

Alright, let's get our hands dirty with some synthetic division! This is a powerful technique that will help us break down the polynomial $2x^4 - 7x^3 - 27x^2 + 63x + 81$. We know that $x = 3$ and $x = -1$ are roots, so we'll use synthetic division to divide the polynomial by $(x - 3)$ and $(x + 1)$. Remember, synthetic division is a streamlined way to divide a polynomial by a linear factor, and it's super efficient. Think of it as the fast track to factorization!

First, let's divide by $(x - 3)$. Set up the synthetic division table with the coefficients of the polynomial (2, -7, -27, 63, 81) and the root 3. Bring down the first coefficient (2), multiply it by the root (3), and add the result to the next coefficient (-7). Continue this process until you reach the end. If 3 is indeed a root, the remainder should be zero. Let's see how it goes:

3 | 2 -7 -27 63 81
 | 6 -3 -90 -81
 ------------------------
 2 -1 -30 -27 0

Look at that! The remainder is 0, confirming that $x = 3$ is a root and $(x - 3)$ is a factor. The resulting coefficients (2, -1, -30, -27) represent the quotient polynomial $2x^3 - x^2 - 30x - 27$. We've successfully reduced the degree of the polynomial by one!

Now, let’s divide this quotient polynomial by $(x + 1)$. We use the same process, but this time with the root -1 and the coefficients 2, -1, -30, -27. This will help us see if -1 is also a root of the quotient polynomial, and further simplify the expression.

-1 | 2 -1 -30 -27
 | -2 3 27
 ------------------
 2 -3 -27 0

Awesome! The remainder is again 0, so $x = -1$ is a root, and $(x + 1)$ is a factor of the quotient polynomial. The new quotient polynomial is $2x^2 - 3x - 27$. We've now gone from a quartic polynomial to a quadratic polynomial – which is much easier to handle. Think of synthetic division as peeling away the layers of the polynomial, revealing its simpler components. Each successful division brings us closer to the linear factors we're aiming for. Keep going guys, we're making great progress!

Factoring the Quadratic

Okay, we've used synthetic division to reduce our quartic polynomial $2x^4 - 7x^3 - 27x^2 + 63x + 81$ down to the quadratic $2x^2 - 3x - 27$. Now, it's time to tackle this quadratic and break it down into linear factors. Factoring quadratics is a fundamental skill in algebra, and there are several ways we can approach it. We can try factoring by grouping, using the quadratic formula, or even just spotting the factors by inspection. Let's see which method works best for us.

One common technique is to look for two numbers that multiply to give the product of the leading coefficient (2) and the constant term (-27), which is -54, and add up to the middle coefficient (-3). This might sound like a mouthful, but it's a systematic way to find the right factors. So, we need two numbers that multiply to -54 and add to -3. After a bit of thinking, we can see that the numbers -9 and 6 fit the bill. Eureka! We've found our magic numbers.

Now, we can rewrite the middle term of the quadratic using these numbers: $2x^2 - 3x - 27 = 2x^2 - 9x + 6x - 27$. This might seem like we're making things more complicated, but it sets us up perfectly for factoring by grouping. We group the first two terms and the last two terms: $(2x^2 - 9x) + (6x - 27)$. Now, we factor out the greatest common factor from each group.

From the first group, we can factor out an $x$, giving us $x(2x - 9)$. From the second group, we can factor out a 3, giving us $3(2x - 9)$. Notice that we now have a common factor of $(2x - 9)$ in both terms! This is the key to factoring by grouping. We factor out the $(2x - 9)$, and we're left with $(x + 3)$. So, our factored quadratic is $(2x - 9)(x + 3)$. Isn't it satisfying when it all comes together?

Alternatively, we could have used the quadratic formula to find the roots of the quadratic equation $2x^2 - 3x - 27 = 0$. The quadratic formula is a reliable method that always works, even when factoring by grouping seems tricky. The roots would give us the factors directly. Or, if you're feeling particularly sharp, you might have recognized the factors just by looking at the quadratic – practice makes perfect!

Final Assembly and Solution

Alright, guys, we've done the heavy lifting! We've used synthetic division to break down the original polynomial and factored the resulting quadratic. Now, it's time to put all the pieces together and express the polynomial as a product of linear factors. This is the moment of truth where everything we've done comes together to give us the final answer. Get ready for some mathematical satisfaction!

We started with the polynomial $2x^4 - 7x^3 - 27x^2 + 63x + 81$ and the goal of writing it in the form $(x + ?)(x - 3)(x + 1)(2x - [])$. Through synthetic division, we confirmed that $(x - 3)$ and $(x + 1)$ are factors. We then reduced the polynomial to the quadratic $2x^2 - 3x - 27$, which we factored as $(2x - 9)(x + 3)$.

So, the complete factorization is $(x - 3)(x + 1)(2x - 9)(x + 3)$. Now, let’s compare this to the target form $(x + ?)(x - 3)(x + 1)(2x - [])$. We can see that the missing constant in the first factor is 3, so $(x + ?) = (x + 3)$. The missing constant in the last factor is 9, so $(2x - []) = (2x - 9)$.

Therefore, the polynomial $2x^4 - 7x^3 - 27x^2 + 63x + 81$ can be written as $(x + 3)(x - 3)(x + 1)(2x - 9)$. We did it! We've successfully expressed the polynomial as a product of linear factors. This is a fantastic achievement, and it demonstrates our understanding of polynomial factorization, synthetic division, and quadratic factoring.

In conclusion, by using a combination of synthetic division and factoring techniques, we were able to break down a complex quartic polynomial into its linear factors. This process not only gives us the factored form but also reveals the roots of the polynomial, which are crucial for many applications in mathematics and beyond. Great job, everyone! Keep practicing, and you'll become a polynomial factorization pro in no time!

Final Answer:

The final answer is $(x+3)(x-3)(x+1)(2x-9)$