Simplify $(3+\sqrt{7})(3-\sqrt{7})$

by Andrew McMorgan 36 views

Hey guys, welcome back to Plastik Magazine! Today, we're diving deep into the fascinating world of mathematics, specifically tackling a problem that might look a bit intimidating at first glance, but trust me, it's a piece of cake once you know the trick. We're going to simplify the product of (3+7)(3−7)(3+\sqrt{7})(3-\sqrt{7}). This is a classic example of a special algebraic identity that makes simplifying these kinds of expressions super straightforward. So, grab your calculators (or just your brains!), and let's get this mathematical party started! We'll break down each step, explore why the options provided are what they are, and ultimately find the correct answer. Get ready to flex those math muscles, because this is going to be fun!

Understanding the Expression

Alright, let's first get a good look at the expression we're working with: (3+7)(3−7)(3+\sqrt{7})(3-\sqrt{7}). The first thing you might notice is that these two binomials look almost identical, differing only by the sign in the middle. This, my friends, is a huge clue! This pattern is known as the difference of squares, and it follows a very handy formula: (a+b)(a−b)=a2−b2(a+b)(a-b) = a^2 - b^2. In our case, aa is 3 and bb is 7\sqrt{7}. Recognizing this pattern is key to solving this problem efficiently and avoiding unnecessary calculations. If you don't immediately spot this, don't worry! We can also solve it using the distributive property, often called FOIL (First, Outer, Inner, Last). Let's see how that works out. Remember, the goal is to find the product, which means we need to multiply these two expressions together. So, let's keep our eyes peeled for that difference of squares pattern because it's going to save us a lot of time and potential mistakes.

Applying the Difference of Squares Identity

Now, let's apply the difference of squares identity to our problem (3+7)(3−7)(3+\sqrt{7})(3-\sqrt{7}). As we identified, a=3a = 3 and b=7b = \sqrt{7}. Plugging these into the formula (a+b)(a−b)=a2−b2(a+b)(a-b) = a^2 - b^2, we get: 32−(7)23^2 - (\sqrt{7})^2. Calculating these squares is pretty simple. 323^2 is just 3×33 \times 3, which equals 9. And (7)2(\sqrt{7})^2 is simply 7, because squaring a square root cancels out the square root operation. So, our expression simplifies to 9−79 - 7. And what is 9−79 - 7, guys? It's 2! So, the product of (3+7)(3−7)(3+\sqrt{7})(3-\sqrt{7}) is 2. This is the most direct and elegant way to solve this. It’s crucial to remember these algebraic identities because they pop up all the time in math, and knowing them can make complex problems feel much simpler. This identity is a real game-changer, turning a potentially messy multiplication into a clean subtraction. Stick around, and we'll also explore the FOIL method to show you how it leads to the same result, reinforcing why the difference of squares is so powerful.

Using the FOIL Method (For Deeper Understanding)

For those of you who didn't spot the difference of squares immediately, or just want to see how it works out using a more fundamental method, let's use the FOIL method to multiply (3+7)(3−7)(3+\sqrt{7})(3-\sqrt{7}). FOIL stands for First, Outer, Inner, Last, and it's a systematic way to ensure you multiply every term in the first binomial by every term in the second.

  • First: Multiply the first terms of each binomial: 3×3=93 \times 3 = 9.
  • Outer: Multiply the outer terms: 3×(−7)=−373 \times (-\sqrt{7}) = -3\sqrt{7}.
  • Inner: Multiply the inner terms: 7×3=37\sqrt{7} \times 3 = 3\sqrt{7}.
  • Last: Multiply the last terms: 7×(−7)=−(7)2=−7\sqrt{7} \times (-\sqrt{7}) = -(\sqrt{7})^2 = -7.

Now, we add all these results together: 9−37+37−79 - 3\sqrt{7} + 3\sqrt{7} - 7. Look at that! We have a −37-3\sqrt{7} and a +37+3\sqrt{7}. These are additive inverses, meaning they cancel each other out. So, −37+37=0-3\sqrt{7} + 3\sqrt{7} = 0. This leaves us with 9−79 - 7, which, as we saw before, equals 2. See? Even without using the specific difference of squares formula, we arrive at the same answer. This method also gives us a clearer view of why the middle terms cancel out in difference of squares problems. It's all about the properties of numbers and how they interact when multiplied. This reinforces the power of algebraic identities as shortcuts, but understanding the underlying mechanics is just as important for building a solid math foundation.

Analyzing the Given Options

Now, let's take a look at the options provided in the question and see how they relate to our calculations. The options are:

A. 6−3ope7+3ope7−ope146-3 ope{7}+3 ope{7}- ope{14} B. 9−ope21+ope21−ope499- ope{21}+ ope{21}- ope{49} C. 9−3ope7+3ope7−ope499-3 ope{7}+3 ope{7}- ope{49}

Let's break each one down. First, option A: 6−3ope7+3ope7−ope146-3 ope{7}+3 ope{7}- ope{14}. This doesn't look like it came from our FOIL expansion at all. The '6' at the beginning is suspicious, and the −ope14- ope{14} at the end doesn't fit. It seems like a mix-up of terms, possibly from trying to multiply numbers that weren't part of the original expression.

Next, option B: 9−ope21+ope21−ope499- ope{21}+ ope{21}- ope{49}. This one starts with a '9', which is good since 3×3=93 \times 3 = 9. However, we have \rope21\rope{21} terms. Where would \rope21\rope{21} come from? It looks like someone might have tried to multiply 7\sqrt{7} by 3, getting 21\sqrt{21}, but then made errors in subsequent steps or perhaps tried to apply FOIL incorrectly with different numbers. Also, −ope49- ope{49} simplifies to -7, which is part of the correct answer, but the \rope21\rope{21} terms canceling out doesn't account for the initial '9' in the context of difference of squares.

Finally, option C: 9−3ope7+3ope7−ope499-3 ope{7}+3 ope{7}- ope{49}. This looks very familiar! This is exactly what we got when we applied the FOIL method before simplifying the last term: 9−37+37−79 - 3\sqrt{7} + 3\sqrt{7} - 7. The only difference is that option C has −ope49- ope{49} instead of −7-7. We know that 49=7\sqrt{49} = 7, so −ope49- ope{49} is indeed −7-7. Therefore, option C represents the intermediate step of the FOIL expansion of (3+7)(3−7)(3+\sqrt{7})(3-\sqrt{7}) before the final simplification.

While the ultimate product is 2, the question asks to choose the product, and option C shows the expanded form derived from FOIL which correctly represents the multiplication before simplification. It’s important to read the question carefully. If the question was