How To Expand (x-9)^2 Easily

Let's dive into the world of algebra and tackle the expression $(x-9)^2$. Expanding algebraic expressions is a fundamental skill, and understanding how to do it will unlock many doors in your mathematical journey. We'll break down $(x-9)^2$ step-by-step, making it super clear and easy to grasp. Get ready to boost your math confidence!

Understanding the Expression: $(x-9)^2$

Before we begin expanding, it's crucial to understand what $(x-9)^2$ actually means. The little '2' above the parentheses tells us we need to multiply the entire expression inside the parentheses by itself. So, $(x-9)^2$ is the same as writing $(x-9) \times (x-9)$. Think of it like squaring a number; for example, $5^2$ means $5 \times 5$. Here, our 'number' is the algebraic expression $(x-9)$.

Why is expanding important? Expanding expressions helps us simplify them, solve equations, and work with more complex algebraic problems. It's like taking a puzzle apart to see all the individual pieces. In this case, we're transforming a compact, squared form into a more spread-out polynomial.

The distributive property: Our best friend The key to expanding expressions like this is the distributive property. This property states that for any numbers or expressions a, b, and c, $a(b+c) = ab + ac$. We'll be using a slightly more extended version of this when multiplying two binomials (expressions with two terms).

Setting the stage for expansion So, our mission is to multiply $(x-9)$ by $(x-9)$. We need to ensure that every term in the first set of parentheses is multiplied by every term in the second set of parentheses. This might sound a bit daunting, but we'll use a systematic approach to make sure we don't miss anything. Let's get started on the actual expansion process, and you'll see just how straightforward it is.

Method 1: The FOIL Method

The FOIL method is a popular and effective way to expand expressions that involve multiplying two binomials. FOIL is an acronym that stands for First, Outer, Inner, Last. It's a mnemonic device to help you remember which pairs of terms to multiply.

Let's apply FOIL to our expression $(x-9)(x-9)$:

1. F (First): Multiply the first term in each binomial. In our case, the first term in the first binomial is '$x$', and the first term in the second binomial is also '$x$'. So, we multiply $x \times x$. Result: $x^2$

2. O (Outer): Multiply the outer terms. These are the terms at the very beginning and the very end of the two binomials. Here, it's '$x$' from the first binomial and '$-9$' from the second binomial. Remember to include the signs! So, we multiply $x \times (-9)$. Result: $-9x$

3. I (Inner): Multiply the inner terms. These are the two terms in the middle. Here, it's '$-9$' from the first binomial and '$x$' from the second binomial. Again, mind the signs. So, we multiply $(-9) \times x$. Result: $-9x$

4. L (Last): Multiply the last term in each binomial. Here, it's '$-9$' from the first binomial and '$-9$' from the second binomial. Two negatives make a positive! So, we multiply $(-9) \times (-9)$. Result: $+81$

Combining the results Now that we have all four products, we add them together: $x^2 + (-9x) + (-9x) + 81$.

Simplifying the expression The next step is to combine like terms. In our result, we have two terms with '$x$' in them: $-9x$ and $-9x$. Combining these gives us $-9x - 9x = -18x$.

So, the fully expanded and simplified expression is $x^2 - 18x + 81$. The FOIL method ensures that we systematically account for all possible multiplications, leading us to the correct expanded form.

Method 2: The Distributive Property (General Approach)

While FOIL is specifically for binomials, the distributive property is a more general concept that underlies all such expansions. Let's see how it works for $(x-9)(x-9)$, reinforcing the FOIL steps but showing the underlying principle.

We want to compute $(x-9)(x-9)$. We can distribute the first binomial $(x-9)$ to each term in the second binomial $(x-9)$.

Think of it as: $(x-9) \times (x \text{ and } -9)$.

So, we distribute the $(x-9)$ to the '$x$' and then distribute the $(x-9)$ to the '$-9$':

1. Distribute $(x-9)$ to '$x$': This gives us $(x-9) \times x$. Using the distributive property again on this part, we get $x \times x - 9 \times x$. Result: $x^2 - 9x$

2. Distribute $(x-9)$ to '$-9$': This gives us $(x-9) \times (-9)$. Using the distributive property here, we get $x \times (-9) - 9 \times (-9)$. Result: $-9x + 81$

Putting it all together Now, we add the results from step 1 and step 2: $(x^2 - 9x) + (-9x + 81)$.

Combining like terms Just like with the FOIL method, we combine the '$x$' terms: $-9x - 9x = -18x$.

The final expanded expression is $x^2 - 18x + 81$. This method shows how the distributive property works sequentially, ensuring every term is accounted for. It's the foundational principle behind FOIL.

Method 3: The Perfect Square Trinomial Formula

For expressions of the form $(a-b)^2$, there's a specific algebraic identity known as the perfect square trinomial formula. This formula provides a shortcut for expanding such expressions directly. The formula states:

$$(a-b)^2 = a^2 - 2ab + b^2$$

Let's apply this formula to our expression $(x-9)^2$. In this case:

• Our '$a$' corresponds to '$x$'.
• Our '$b$' corresponds to '$9$'.

Now, we substitute these values into the formula:

1. $a^2$: This means $x^2$. Result: $x^2$

2. $-2ab$: This means $-2 \times x \times 9$. When we multiply this out, we get $-18x$. Result: $-18x$

3. $+b^2$: This means $+9^2$. And $9^2$ is $9 \times 9$, which equals $81$. Result: $+81$

Combining the terms Putting these results together according to the formula, we get: $x^2 - 18x + 81$.

Why this formula works This formula is essentially a generalization derived from the FOIL or distributive property methods. If you were to expand $(a-b)(a-b)$ using FOIL, you would get:

• First: $a \times a = a^2$
• Outer: $a \times (-b) = -ab$
• Inner: $(-b) \times a = -ab$
• Last: $(-b) \times (-b) = +b^2$

Adding these gives $a^2 - ab - ab + b^2 = a^2 - 2ab + b^2$. Recognizing this pattern allows for quicker expansion.

When to use this shortcut This formula is particularly useful when you encounter an expression that perfectly fits the $(a-b)^2$ or $(a+b)^2$ format. It saves time and reduces the chance of making errors in multiplication.

Conclusion: Mastering $(x-9)^2$ Expansion

We've explored three effective methods to expand the expression $(x-9)^2$: the FOIL method, the general distributive property, and the specific perfect square trinomial formula. Each method leads to the same correct answer: $x^2 - 18x + 81$. Understanding these different approaches not only helps you solve this particular problem but also builds a strong foundation for tackling more complex algebraic manipulations in the future.

Key takeaways

• $(x-9)^2$ means $(x-9) \times (x-9)$.
• Remember to multiply every term in the first binomial by every term in the second.
• Pay close attention to the signs, especially when multiplying negative numbers.
• Always combine like terms at the end to simplify your answer.

Practice makes perfect

The more you practice expanding expressions, the more comfortable and proficient you'll become. Try expanding similar expressions, like $(x+5)^2$, $(2x-3)^2$, or even $(y+4)(y-4)$. Each practice problem reinforces the concepts and builds your mathematical intuition.

If you're looking to deepen your understanding of algebraic identities and expansions, exploring resources like Khan Academy can provide further explanations, examples, and practice exercises. They offer a wealth of information on topics ranging from basic algebra to advanced calculus, all presented in an accessible way.

Khan Academy is an excellent resource for anyone looking to strengthen their math skills.