Divide Polynomials: (x³ + 8) ÷ (x + 2) Solution

Hey guys! Let's dive into a fun math problem today. We're going to tackle polynomial division, which might sound intimidating, but I promise it's totally manageable. Our mission, should we choose to accept it, is to figure out the quotient of $(x^3 + 8) ÷ (x + 2)$. In simpler terms, we need to divide the polynomial $x^3 + 8$ by the binomial $x + 2$ and see what we get.

Understanding Polynomial Division

Polynomial division, at its core, is very similar to the long division you learned way back in elementary school, but instead of numbers, we're dealing with expressions that contain variables and exponents. Think of it as breaking down a complex expression into smaller, more manageable parts. In this case, we want to see how many times $(x + 2)$ fits into $(x^3 + 8)$. Before we jump into solving the problem, let’s quickly recap some key concepts that’ll help us along the way.

First, remember what a quotient is. The quotient is the result you get after dividing one number (or polynomial) by another. It's the answer to our division problem. Next, let's talk about the anatomy of a polynomial. A polynomial is an expression that consists of variables (like x) raised to non-negative integer powers, combined with constants (numbers). Our dividend, $x^3 + 8$, is a polynomial. Notice something interesting about it? It seems to be missing some terms. We have an $x^3$ term and a constant term (8), but where are the $x^2$ and $x$ terms? This is where a little trick comes in handy: we can rewrite the polynomial as $x^3 + 0x^2 + 0x + 8$. Adding these zero terms doesn't change the value of the polynomial, but it does help us keep everything organized during the division process.

Why is this important? Well, when we perform polynomial division, we need to make sure we're aligning like terms – that is, terms with the same exponent. Adding the zero terms acts like placeholders, ensuring that our columns line up correctly. Now, let's consider our divisor, which is $(x + 2)$. This is a binomial, meaning it's a polynomial with two terms. It's a relatively simple expression, but it plays a crucial role in our division. The goal of polynomial division is to find another polynomial (the quotient) that, when multiplied by our divisor $(x + 2)$, gives us our original dividend $(x^3 + 8)$. In essence, we're trying to “undo” multiplication.

To successfully tackle polynomial division, it's essential to be comfortable with basic algebraic operations, such as adding, subtracting, multiplying, and dividing terms with exponents. You should also be familiar with the distributive property, which is key to multiplying polynomials. The distributive property states that $a(b + c) = ab + ac$. We'll use this property to multiply our quotient terms by the divisor. Keep in mind that the process of polynomial division is iterative, meaning we repeat certain steps until we arrive at a remainder that has a lower degree than the divisor. It might seem a bit complex at first, but with practice, it becomes much more intuitive. We'll break down each step in detail as we solve our problem, so you can see exactly how it works.

Setting Up the Division

Okay, let's get down to business and set up our division problem. Remember how we do long division with numbers? We'll use a similar setup here. We write the dividend, which is $x^3 + 0x^2 + 0x + 8$, inside the division symbol, and the divisor, $x + 2$, outside. It looks a little something like this:

 x + 2 | x³ + 0x² + 0x + 8

See how we included those zero terms? They're going to be our best friends in keeping everything aligned. Now, the fun begins! The first question we ask ourselves is: what do we need to multiply $x$ (the first term of the divisor) by to get $x^3$ (the first term of the dividend)? The answer is $x^2$. So, we write $x^2$ above the $x^2$ column in the quotient area. This is our first term of the quotient. Next, we multiply this $x^2$ by the entire divisor $(x + 2)$. Using the distributive property, we get $x^2 * (x + 2) = x^3 + 2x^2$. We write this result below the corresponding terms in the dividend:

 x²
 x + 2 | x³ + 0x² + 0x + 8
       x³ + 2x²

Now, we subtract the expression we just wrote from the corresponding terms in the dividend. This is where the zero terms really come in handy, as they help us keep track of our subtractions. Subtracting $(x^3 + 2x^2)$ from $(x^3 + 0x^2)$ gives us:

 (x³ + 0x²) - (x³ + 2x²) = x³ + 0x² - x³ - 2x² = -2x²

We bring down the next term from the dividend, which is $0x$, to join our result. This gives us $-2x^2 + 0x$. Our setup now looks like this:

 x²
 x + 2 | x³ + 0x² + 0x + 8
       x³ + 2x²
       ---------
       -2x² + 0x

We're halfway through the process! We've completed one cycle of dividing, multiplying, and subtracting. Now, we simply repeat these steps until we've brought down all the terms from the dividend.

Continuing the Division Process

Alright, we've made a good start! We're now looking at $-2x^2 + 0x$. The next question we ask ourselves is: what do we need to multiply $x$ (the first term of the divisor) by to get $-2x^2$ (the new first term)? The answer this time is $-2x$. So, we write $-2x$ next to the $x^2$ in the quotient area. Remember, the quotient is the result of our division, so we're building it piece by piece. Now, we multiply $-2x$ by the entire divisor $(x + 2)$. Again, using the distributive property:

 -2x * (x + 2) = -2x² - 4x

We write this result below the corresponding terms:

 x² - 2x
 x + 2 | x³ + 0x² + 0x + 8
       x³ + 2x²
       ---------
       -2x² + 0x
       -2x² - 4x

Time for another subtraction! We subtract $(-2x^2 - 4x)$ from $(-2x^2 + 0x)$:

 (-2x² + 0x) - (-2x² - 4x) = -2x² + 0x + 2x² + 4x = 4x

The $-2x^2$ terms cancel out, leaving us with $4x$. We bring down the last term from the dividend, which is $+8$, to join our result. This gives us $4x + 8$. Our setup now looks like this:

 x² - 2x
 x + 2 | x³ + 0x² + 0x + 8
       x³ + 2x²
       ---------
       -2x² + 0x
       -2x² - 4x
       ---------
       4x + 8

We're in the home stretch now! One more cycle to go. We ask ourselves one last time: what do we need to multiply $x$ (the first term of the divisor) by to get $4x$ (the new first term)? This time, the answer is simply $4$. So, we write $+4$ next to the $-2x$ in the quotient area. We multiply $4$ by the entire divisor $(x + 2)$:

 4 * (x + 2) = 4x + 8

We write this result below the corresponding terms:

 x² - 2x + 4
 x + 2 | x³ + 0x² + 0x + 8
       x³ + 2x²
       ---------
       -2x² + 0x
       -2x² - 4x
       ---------
       4x + 8
       4x + 8

And finally, we subtract $(4x + 8)$ from $(4x + 8)$:

 (4x + 8) - (4x + 8) = 4x + 8 - 4x - 8 = 0

We get a remainder of $0$. This means that $(x + 2)$ divides evenly into $(x^3 + 8)$. Hooray!

The Solution: Interpreting the Quotient

So, what's the quotient we've been working so hard to find? Look at the top of our division setup – it's $x^2 - 2x + 4$. This is the polynomial we get when we divide $(x^3 + 8)$ by $(x + 2)$. In other words:

 (x³ + 8) ÷ (x + 2) = x² - 2x + 4

That's it! We've successfully performed polynomial division and found our answer. This matches option B from our choices. Polynomial division might seem a little daunting at first, but with a bit of practice and careful attention to detail, it's a skill you can definitely master. Remember the key steps: divide, multiply, subtract, and bring down. Keep practicing, and you'll be dividing polynomials like a pro in no time!

Why This Matters: Applications of Polynomial Division

You might be wondering,