We Distribute, Yet Things Multiply

Namaste, dear students! Welcome to today's mathematics lesson. I am so happy to see you all here, ready to learn something new and exciting. Today, we are going to study Chapter 6, and the title is so interesting that it makes you think, doesn't it? "We Distribute, Yet Things Multiply." What does this mean? Well, by the end of this chapter, you will not only understand this title but also appreciate the beauty of a very important property in mathematics called the distributive property. So let's begin our journey together.

In our previous chapters, we have seen how algebra uses letter symbols to write general statements about patterns and relations in a compact manner. We have also seen how algebra can be used to justify or prove claims and to solve problems of various kinds. Today, we will explore one such property that is extremely powerful and useful - the distributive property of multiplication over addition.

Let us start with a very simple question. Consider the multiplication of two numbers, say 23 multiplied by 27. Now, I want you to think about what happens to this product if we change one of the numbers. Specifically, I want you to consider these three questions:

First, by how much does the product increase if the first number, which is 23, is increased by 1? Second, what if the second number, which is 27, is increased by 1? And third, how about when both numbers are increased by 1? Do you see a pattern that could help us generalize our observations to the product of any two numbers?

Let us first consider the simpler problem - find the increase in the product when 27 is increased by 1. From the definition of multiplication and the commutative property, it is clear that the product increases by 23. But let us see how this works using the distributive property. If a, b, and c are three numbers, then we have a multiplied by the quantity b plus c equals a multiplied by b plus a multiplied by c. In mathematical notation, this is a(b + c) = ab + ac. This is what we call the distributive property of multiplication over addition.

Let me explain this with our example. We have 23 multiplied by 27. If we increase 27 by 1, we get 23 multiplied by the quantity 27 plus 1. Using the distributive property, this equals 23 multiplied by 27 plus 23 multiplied by 1, which is 23 × 27 + 23. So the increase is exactly 23, as we expected. Notice that here a(b + c) and 23(27 + 1) mean a × (b + c) and 23 × (27 + 1) respectively. We usually skip writing the multiplication symbol before or after brackets, just as we do in expressions like 5a or xy.

Now, let us also see how we can expand (a + b)c using the distributive property. We have (a + b)c = c(a + b) by commutativity of multiplication, which equals ca + cb by distributivity, and this equals ac + bc again by commutativity. So you see, the distributive property works in both directions.

Now, let us use this property to find out how much a product increases if one or both the numbers in the product are increased by 1. Suppose the initial two numbers are a and b. If one of the numbers, say b, is increased by 1, then we have a multiplied by the quantity b plus 1, which equals ab + a by the distributive property. So the increase is a.

Now let us see what happens if both numbers in a product are increased by 1. If in the product ab, both a and b are increased by 1, then we obtain the quantity a plus 1 multiplied by the quantity b plus 1. How do we expand this?

Let us consider (a + 1) as a single term. Then, by the distributive property, we have (a + 1)(b + 1) = (a + 1)b + (a + 1) × 1. Now, if we take a = 23 and b = 27, we get (23 + 1)(27 + 1) = (23 + 1) × 27 + (23 + 1) × 1. But let us do this more generally.

Again applying the distributive property, we obtain (a + 1)(b + 1) = (a + 1)b + (a + 1) = ab + b + a + 1. So the increase is a + b + 1. Let me write this clearly: when both a and b are increased by 1, the product ab increases by a + b + 1.

Let me verify this with our example. If a = 23 and b = 27, then the original product is 23 × 27. After increasing both by 1, we get 24 × 28. The increase should be 23 + 27 + 1 = 51. Let us check: 24 × 28 = 672, and 23 × 27 = 621. The difference is indeed 51. So our formula works!

Now, what would we get if we had expanded (a + 1)(b + 1) by first taking (b + 1) as a single term? I would like you to try this on your own. You will get the same result, which is ab + a + b + 1.

Now, here is another interesting question. What happens when one of the numbers in a product is increased by 1 and the other is decreased by 1? Will there be any change in the product? Let us find out.

Let us take the product ab of two numbers a and b. If a is increased by 1 and b is decreased by 1, then their product will be (a + 1)(b - 1). Expanding this, we get (a + 1)(b - 1) = (a + 1)b - (a + 1) × 1 = ab + b - a - 1. So the change in the product is b - a - 1. If we take a = 23 and b = 27, then the change is 27 - 23 - 1 = 3. Let us verify: (23 + 1)(27 - 1) = 24 × 26 = 624, and 23 × 27 = 621. The difference is indeed 3. But wait, is this an increase or a decrease? Since the result is positive, the product increases by 3.

Now, I want you to think about this: will the product always increase when one number is increased and the other is decreased? Find 3 examples where the product decreases. This is an interesting exercise that will help you understand the behavior of products when numbers change.

What happens when a and b are negative integers? We should check this as well. For example, take a = -5 and b = 8. Then (a + 1)(b - 1) = (-5 + 1)(8 - 1) = (-4)(7) = -28, and ab = (-5)(8) = -40. The difference is -28 - (-40) = 12, which is an increase. Try some other values yourself, including cases where both a and b are negative.

We have seen that integers also satisfy the distributive property. That is, if x, y, and z are any three integers, then x(y + z) = xy + xz. Thus, the expressions we have derived for the increase of products hold when the letter-numbers take on negative integer values as well.

Now, let me introduce you to a very important concept. We know that two algebraic expressions are equal if they take on the same values when their letter-numbers are replaced by numbers. These numbers could be any integers. Mathematical statements that express the equality of two algebraic expressions, such as a(b + 8) = ab + 8a, or (a + 1)(b - 1) = ab + b - a - 1, are called identities. An identity is a statement that is true for all values of the variables involved. We will be seeing many such identities in this chapter.

Now, let us generalize further. By how much will the product of two numbers change if one of the numbers is increased by m and the other by n? If a and b are the initial numbers being multiplied, they become a + m and b + n. Let us expand (a + m)(b + n) using the distributive property.

We have (a + m)(b + n) = (a + m)b + (a + m)n = ab + mb + an + mn. So the increase is an + bm + mn. Notice that the product is the sum of the product of each term of (a + m) with each term of (b + n). This is a very important observation, and it gives us our first major identity:

Identity 1: (a + m)(b + n) = ab + mb + an + mn

This identity can be used to find how products change when the numbers being multiplied are increased or decreased by any amount. Can you see how this identity can be used when one or both numbers are decreased? For example, if we want to decrease one number, we can take m or n to be negative.

For instance, let us reconsider the case when one number is increased by 1 and the other decreased by 1. Let us write the product (a + 1)(b - 1) as (a + 1)(b + (-1)). Taking m = 1 and n = -1 in Identity 1, we have ab + (1) × b + a × (-1) + (1) × (-1) = ab + b - a - 1, which is the same expression that we obtained earlier.

Now, let us use Identity 1 to find how the product changes when one number is decreased by 2 and the other increased by 3. In this case, m = -2 and n = 3. The increase would be a × 3 + b × (-2) + (-2) × 3 = 3a - 2b - 6. Similarly, if both numbers are decreased, one by 3 and the other by 4, then m = -3 and n = -4, and the increase would be a × (-4) + b × (-3) + (-3) × (-4) = -4a - 3b + 12. You can verify these answers by finding the products without converting the subtractions to additions.

Generalizing this further, we can find the product (a + u)(b - v) as follows. (a + u)(b - v) = (a + u)b - (a + u)v = ab + ub - (av + uv) = ab + ub - av - uv. Check that this is the same as taking m = u and n = -v in Identity 1.

As in Identity 1, the product (a + u)(b - v) is the sum of the product of each term of a + u (that is, a and u) with each term of b - v (that is, b and -v). Notice that the signs of the terms in the products can be determined using the usual rules of integer multiplication. This is very convenient because we can handle multiple cases using a single identity!

Now, let me give you some exercises to practice. Expand (a - u)(b + v) and (a - u)(b - v). We get (a - u)(b + v) = ab - ub + av - uv, and (a - u)(b - v) = ab - ub - av + uv. Notice how the signs change depending on whether we are adding or subtracting.

The distributive property is not restricted to two terms within a bracket. Let me show you an example.

Example 1: Expand (3a/2)(a - b + 1/5).

We have (3a/2)(a - b + 1/5) = (3a/2 × a) - (3a/2 × b) + (3a/2 × 1/5).

The terms can be simplified as follows: 3a/2 × a = 3/2 × (a × a). Using exponent notation, we can write this as 3/2 a². Next, 3a/2 × b = 3/2 × (a × b) = 3/2 ab. And 3a/2 × 1/5 = (3/2 × 1/5)a = 3/10 a.

So we get (3a/2)(a - b + 1/5) = 3/2 a² - 3/2 ab + 3/10 a.

Now, can any two terms be added to get a single term? For example, can 3/2 a² and 3/10 a be added? We see that no two terms have exactly the same letter-numbers, which would have allowed them to be simplified into a single term. So, a further simplification of the expression is not possible. Recall that we call terms having the same letter-numbers like terms.

Example 2: Expand (a + b)(a + b).

We have (a + b)(a + b) = (a + b)a + (a + b)b = a × a + b × a + a × b + b × b = a² + ba + ab + b².

Since ba = ab, we have two terms having the same letter-numbers ab, or in other words, like terms, and so they can be added. ba + ab = ab + ab = 2ab.

So we get (a + b)(a + b) = a² + 2ab + b².

This is a very important identity, and we will be using it frequently. Notice that this is the square of the sum of two numbers.

Example 3: Expand (a + b)(a² + 2ab + b²).

We have (a + b)(a² + 2ab + b²) = (a + b)a² + (a + b) × 2ab + (a + b)b² = (a × a²) + ba² + (a × 2ab) + (b × 2ab) + ab² + (b × b²).

The terms can be simplified as follows: a × a² = a³ (because a multiplied by a² gives a³). ba² = a²b. a × 2ab = 2 × a × a × b = 2a²b. b × 2ab = 2 × a × b × b = 2ab². b × b² = b³.

So, (a + b)(a² + 2ab + b²) = a³ + a²b + 2a²b + 2ab² + ab² + b³.

We see that a²b and 2a²b have the same letter-numbers, or are like terms, and so can be added: a²b + 2a²b = (1 + 2)a²b = 3a²b. Similarly, ab² and 2ab² are like terms and so can be added: ab² + 2ab² = (1 + 2)ab² = 3ab².

Thus, we have (a + b)(a² + 2ab + b²) = a³ + 3a²b + 3ab² + b³.

This is another very important identity. You will notice that it follows a pattern similar to the expansion of (a + b)², but with higher powers.

Now, before we move on, let me tell you a bit about the history of the distributive property. The distributive property of multiplication over addition was implicit in the calculations of mathematicians in many ancient civilizations, particularly in ancient Egypt, Mesopotamia, Greece, China, and India. For example, the mathematicians Euclid in geometric form and Āryabhata in algebraic form used the distributive law in an implicit manner extensively in their mathematical and scientific works. The first explicit statement of the distributive property was given by Brahmagupta in his work Brahmasphuṭasiddhānta, which was written around 628 CE. He referred to the use of the property for multiplication as khanda-gunanam, which means multiplication by parts. His verse states, "The multiplier is broken up into two or more parts whose sum is equal to it; the multiplicand is then multiplied by each of these and the results added." That is, if there are two parts, then using letter symbols this is equivalent to the identity (a + b)c = ac + bc. In the next verse, Brahmagupta further describes a method for doing fast multiplication using this distributive property, which we will explore in the next section.

Now, let us learn how to use the distributive property for fast multiplication when certain types of numbers are multiplied. This is especially useful when one of the numbers is 11, 101, 1001, and so on.

Let us first look at how to multiply a number by 11. Consider the multiplication 3874 × 11. We can write 3874 × 11 as 3874 × (10 + 1) = 3874 × 10 + 3874. This equals 38740 + 3874. Now, if we add these, we get 42614. Notice how the digits are getting added in a particular pattern.

Let us take a 4-digit number dcba, that is, the number that has d in the thousands place, c in the hundreds place, b in the tens place, and a in the units place. Then dcba × 11 = dcba × (10 + 1) = dcba × 10 + dcba. This becomes:

d c b a 0 + d c b a ----------- d (c+d) (b+c) (a+b) a

This can be used to obtain the product in one line. So the general rule to multiply a number by 11 is: write the digits, add each digit to its neighbor to the left, and carry over if needed. The rightmost digit stays as it is, the next digit is the sum of the last two digits, and so on.

For example, to multiply 3874 by 11, we start from the right: the rightmost digit is 4. Then we add 4 and 7 to get 11, so we write 1 and carry 1. Then we add 8 and 7 plus the carry 1 to get 16, so we write 6 and carry 1. Then we add 3 and 8 plus the carry 1 to get 12, so we write 2 and carry 1. Finally, we add the leftmost digit 3 and the carry 1 to get 4. So the answer is 42614.

Now, can we come up with a similar rule for multiplying a number by 101? Let us try multiplying 3874 by 101. We can write 3874 × 101 as 3874 × (100 + 1) = 3874 × 100 + 3874. This equals 387400 + 3874 = 391274.

Let us take a 4-digit number dcba. Then dcba × 101 = dcba × (100 + 1) = dcba × 100 + dcba. This becomes:

d c b a 0 0 + d c b a ------------- d c (b+d) (a+c) b a

So the rule for multiplying by 101 is: the first two digits remain the same, then we add each digit to the digit two places to its right, and the last two digits remain the same.

We can extend this rule for multiplication by 1001, 10001, and so on. For 1001, we would add each digit to the digit three places to its right. For 10001, we would add each digit to the digit four places to its right, and so on.

Now, let us also see how to multiply by 99 or 999. Notice that 99 = 100 - 1, and 999 = 1000 - 1. So we can use the distributive property to multiply by these numbers as well. For example, 9734 × 99 = 9734 × (100 - 1) = 9734 × 100 - 9734 × 1 = 973400 - 9734 = 963666. Similarly, 23478 × 999 = 23478 × (1000 - 1) = 23478 × 1000 - 23478 × 1 = 23478000 - 23478 = 23454522.

Now, let us move on to a very important section - the special cases of the distributive property. These are the square of the sum or difference of two numbers.

Consider the area of a square of side length 60 units, which is 3600 square units (that is, 60²), and that of a square of side length 5 units, which is 25 square units (that is, 5²). Can we use this to find the area of a square of side length 65 units?

A square of side length 65 can be split into 4 regions - a square of side length 60, a square of side length 5, and two rectangles of side lengths 60 and 5. The area of the square of side length 65 is the sum of the areas of all its constituent parts.

So we get 65² = (60 + 5)² = 60² + 5² + 2 × (60 × 5) = 3600 + 25 + 600 = 4225 square units.

Let us multiply (60 + 5) × (60 + 5) using the distributive property. (60 + 5) × (60 + 5) = 60 × 60 + 5 × 60 + 60 × 5 + 5 × 5 = 60² + 2 × (60 × 5) + 5².

What if we write 65² as (30 + 35)² or (52 + 13)²? We would get the same answer. You can draw the figures and check the area that you get.

Now, let us look at the general expression for the square of the sum of two numbers, (a + b)². Using the distributive property, (a + b)² can be expanded as (a + b) × (a + b) = a × a + a × b + b × a + b × b = a² + 2ab + b², as we had already seen in Example 2.

This is Identity 1A: (a + b)² = a² + 2ab + b².

Now, I want you to think about this: if a and b are any two integers, is (a + b)² always greater than a² + b²? If not, when is it greater? We will explore this later.

We can use Identity 1A to find the values of squares like 104² and 37². The hint is to decompose 104 and 37 into sums or differences of numbers whose squares are easy to compute. For example, 104 = 100 + 4, so 104² = (100 + 4)² = 100² + 2 × 100 × 4 + 4² = 10000 + 800 + 16 = 10816. Similarly, 37 = 40 - 3, so 37² = (40 - 3)² = 40² - 2 × 40 × 3 + 3² = 1600 - 240 + 9 = 1369.

We can also use Identity 1A to write the expressions for (m + 3)² and (6 + p)². For (m + 3)², we have m² + 2 × m × 3 + 3² = m² + 6m + 9. For (6 + p)², we have 6² + 2 × 6 × p + p² = 36 + 12p + p².

Now, let us expand (6x + 5)². We can do this using the distributive property or using the identity. Using the distributive property: (6x + 5)² = (6x + 5)(6x + 5) = (6x × 6x) + (5 × 6x) + (6x × 5) + 5 × 5 = (6x)² + 2(6x × 5) + 5² = 36x² + 60x + 25. Using the identity: (6x + 5)² = (6x)² + 5² + 2 × (6x × 5) = 36x² + 25 + 60x. Both give the same answer, just written in a different order.

Now, can we use 60² (= 3600) and 5² (= 25) to find the value of (60 - 5)² or 55²? Let us approach this through geometry by drawing a square of side length 55 sitting inside a square of side length 60.

The area of a square of side length 55 is (60 - 5)² = 55². We can get the area of the square of side length 55 by taking the area of the square of side length 60 and removing the areas of the two rectangles of side lengths 60 and 5, that is, 60² - (60 × 5) - (5 × 60). By doing this, we remove the area of the small square of side length 5 twice. What can we do with this expression to get the actual area? We can add back the area of the square of side length 5 to this expression. That way, we are only subtracting this area once.

So, (60 - 5)² = 60² - (60 × 5) - (5 × 60) + 5² = 3600 - 300 - 300 + 25 = 3025. The area of the square of side length 55 is 3025 square units.

Now, what is the expansion of (a - b)²? Using the distributive property, (a - b)² = (a - b) × (a - b) = a² - ba - ab + b² = a² - 2ab + b².

We can also use the expansion of (a + b)² to find the expansion of (a - b)². How? We can think of (a - b)² as (a + (-b))². Then, using the expansion of (a + b)² with b replaced by -b, we get (a + (-b))² = a² + (-b)² + 2 × a × (-b) = a² + b² - 2ab.

This is Identity 1B: (a - b)² = a² + b² - 2ab.

We can use the identity (a - b)² to find the values of 99² and 58². For 99², we can write 99 = 100 - 1, so 99² = (100 - 1)² = 100² - 2 × 100 × 1 + 1² = 10000 - 200 + 1 = 9801. For 58², we can write 58 = 60 - 2, so 58² = (60 - 2)² = 60² - 2 × 60 × 2 + 2² = 3600 - 240 + 4 = 3364.

Now, let us expand (b - 6)², (-2a + 3)², and (7y - 3/4z)² using both Identity 1B and by applying the distributive property.

For (b - 6)², using the identity: b² + 6² - 2 × b × 6 = b² + 36 - 12b. Using the distributive property: (b - 6)(b - 6) = b² - 6b - 6b + 36 = b² - 12b + 36.

For (-2a + 3)², we can write this as (3 - 2a)². Using the identity: (3)² + (-2a)² - 2 × 3 × (-2a) = 9 + 4a² + 12a = 4a² + 12a + 9. Using the distributive property: (-2a + 3)(-2a + 3) = 4a² - 6a - 6a + 9 = 4a² - 12a + 9. Wait, there seems to be a difference. Let me check: -2a + 3 means we have -2a and +3. So (-2a + 3)² = (-2a)² + 3² + 2 × (-2a) × 3 = 4a² + 9 - 12a = 4a² - 12a + 9. Yes, that's correct.

For (7y - 3/4z)², using the identity: (7y)² + (3/4z)² - 2 × 7y × (3/4z) = 49y² + 9/16 z² - (42/4)yz = 49y² + 9/16 z² - 10.5yz. Actually, let me recalculate: 2 × 7y × (3/4z) = 2 × 7 × 3/4 × yz = (42/4)yz = 10.5yz. But it's better to keep it as a fraction: 2 × 7y × (3/4z) = (14y) × (3/4z) = (42/4)yz = (21/2)yz. So we have 49y² + 9/16 z² - 21/2 yz. Using the distributive property would give the same result.

Now, let us investigate some patterns. This is where algebra helps us understand and prove patterns that we observe with numbers.

Pattern 1: Look at the following pattern. 2(2² + 1²) = 3² + 1² 2(3² + 1²) = 4² + 2² 2(6² + 5²) = 11² + 1² 2(5² + 3²) = 8² + 2²

Take a pair of natural numbers. Calculate the sum of their squares. Can you write twice this sum as a sum of two squares? Try this with other pairs of numbers. Have you figured out a pattern?

Notice that 2(5² + 6²) = (6 + 5)² + (6 - 5)².

Do the identities below help in explaining the observed pattern?

(a + b)² = a² + 2ab + b² (a - b)² = a² - 2ab + b²

If we add these two identities, we get (a + b)² + (a - b)² = (a² + 2ab + b²) + (a² - 2ab + b²). Adding the like terms, a² + a² = 2a², b² + b² = 2b², and 2ab - 2ab = 0, we get 2(a² + b²) = (a + b)² + (a - b)².

This is a very beautiful identity! It tells us that twice the sum of squares of two numbers equals the sum of the squares of their sum and their difference.

Pattern 2: Here is a related pattern. Try to describe the pattern using algebra to determine if the pattern always holds.

9 × 9 - 1 × 1 = 10 × 8 8 × 8 - 6 × 6 = 14 × 2 7 × 7 - 2 × 2 = 9 × 5 10 × 10 - 4 × 4 = 14 × 6

The pattern here appears to be a² - b² = (a + b) × (a - b).

Is this a true identity? Using the distributive property, we get (a + b) × (a - b) = a² - ab + ba - b². Adding the like terms, ab + (-ab) = 0, we see that indeed (a + b)(a - b) = a² - b².

This is Identity 1C: (a + b)(a - b) = a² - b².

We can use Identity 1C to calculate 98 × 102 and 45 × 55. Notice that 98 = 100 - 2 and 102 = 100 + 2, so 98 × 102 = (100 - 2)(100 + 2) = 100² - 2² = 10000 - 4 = 9996. Similarly, 45 × 55 = (50 - 5)(50 + 5) = 50² - 5² = 2500 - 25 = 2475.

We can also show that (a + b)(a - b) = a² - b² geometrically. You can try to draw a figure to see this.

Now, Sridharacharya, who lived around 750 CE, gave an interesting method to quickly compute the squares of numbers using Identity 1C. Consider the following modified form of this identity: a² = (a + b)(a - b) + b². Why is this identity true? Well, from Identity 1C, we have (a + b)(a - b) = a² - b², so adding b² to both sides gives a² = (a + b)(a - b) + b².

Now, for example, 31² can be found by taking a = 31 and b = 1. Then 31² = (31 + 1)(31 - 1) + 1² = 32 × 30 + 1 = 960 + 1 = 961. Similarly, 197² can be found by taking a = 197 and b = 3. Then 197² = (197 + 3)(197 - 3) + 3² = 200 × 194 + 9 = 38800 + 9 = 38809. This is much faster than multiplying 197 by 197 directly!

Now, let us move on to a very interesting section called "Mind the Mistake, Mend the Mistake." In this section, we will look at some algebraic expressions that have been simplified, and we need to check if there are any mistakes.

Let me go through each one with you.

First simplification: -3p(-5p + 2q) = -3p + 5p - 2q = p - 2q. Is this correct? Let us check. -3p multiplied by -5p gives 15p². -3p multiplied by 2q gives -6pq. So the correct answer should be 15p² - 6pq. The given simplification is wrong. What could have gone wrong? Probably, the student forgot to multiply p with -3p and just wrote -3p. The correct expression is 15p² - 6pq.

Second simplification: 2(x - 1) + 3(x + 4) = 2x - 1 + 3x + 4 = 5x + 3. Is this correct? Let us check. 2(x - 1) = 2x - 2, not 2x - 1. So the student made a mistake in expanding 2(x - 1). The correct simplification is 2x - 2 + 3x + 12 = 5x + 10.

Third simplification: y + 2(y + 2) = (y + 2)² = y² + 4y + 4. Is this correct? No, y + 2(y + 2) is not equal to (y + 2)². y + 2(y + 2) = y + 2y + 4 = 3y + 4. The student incorrectly squared the expression. The correct expression is 3y + 4.

Fourth simplification: (5m + 6n)² = 25m² + 36n². Is this correct? No! When we square a binomial, we get three terms, not two. The correct expansion is (5m)² + 2 × 5m × 6n + (6n)² = 25m² + 60mn + 36n².

Fifth simplification: (-q + 2)² = q² - 4q + 4. Is this correct? Let us check. (-q + 2)² = (-q)² + 2² + 2 × (-q) × 2 = q² + 4 - 4q = q² - 4q + 4. Yes, this is correct!

Sixth simplification: 3a(2b × 3c) = 6ab × 9ac = 54a²bc. Is this correct? No! The student made several mistakes. First, 2b × 3c = 6bc, not 6ab. Then, 3a × 6bc = 18abc. The correct answer is 18abc.

Seventh simplification: 1/2(10s - 6) + 3 = 5s - 3 + 3 = 5s. Is this correct? Let us check. 1/2 × (10s - 6) = 5s - 3. Then adding 3 gives 5s - 3 + 3 = 5s. Yes, this is correct!

Eighth simplification: 5w² + 6w = 11w². Is this correct? No! These are like terms only if they have the same power of w. 5w² and 6w are not like terms because one has w² and the other has w. They cannot be combined. The expression 5w² + 6w is already in its simplest form.

Ninth simplification: 2a³ + 3a³ + 6a²b + 6ab² = 5a³ + 12a²b². Is this correct? Let us check. 2a³ + 3a³ = 5a³. But 6a²b and 6ab² are not like terms, so they cannot be added. And even if they could be added, 6a²b + 6ab² is not equal to 12a²b². The correct simplification is 5a³ + 6a²b + 6ab².

Tenth simplification: (x + 2)(x + 5) = (x + 2)x + (x + 2)5 = x² + 2x + 5x + 10 = x² + 7x + 10. Is this correct? Yes, this is correct!

Eleventh simplification: (a + 2)(b + 4) = ab + 8. Is this correct? No! The student only multiplied the constant terms and forgot the other terms. The correct expansion is (a + 2)(b + 4) = ab + 4a + 2b + 8.

Twelfth simplification: ab² + a²b + a²b² = ab(a + b + ab). Is this correct? Let us check. ab(a + b + ab) = a²b + ab² + a²b². Yes, this is correct! The student factored out ab correctly.

Now, let us look at a section called "This Way or That Way, All Ways Lead to the Bay." This is about finding different methods to solve the same problem.

Observe the pattern in the figure described. There are circles arranged in a particular way. In Step 1, there are 2 circles. In Step 2, there are 6 circles. In Step 3, there are 12 circles. In Step 4, there are 20 circles. These numbers follow the pattern: 1×2, 2×3, 3×4, 4×5.

We need to find how many circles there are in Step k, and also in Step 10.

There are many ways to interpret this pattern. Let me show you four different methods.

Method 1: The number of circles in Step k is (k + 1)² - (k + 1). This is because Step 1 has (1 + 1)² - (1 + 1) = 4 - 2 = 2 circles, Step 2 has (2 + 1)² - (2 + 1) = 9 - 3 = 6 circles, Step 3 has (3 + 1)² - (3 + 1) = 16 - 4 = 12 circles, and Step 4 has (4 + 1)² - (4 + 1) = 25 - 5 = 20 circles.

Method 2: The number of circles in Step k is k² + k. This gives Step 1: 1 + 1 = 2, Step 2: 4 + 2 = 6, Step 3: 9 + 3 = 12, Step 4: 16 + 4 = 20.

Method 3: The number of circles in Step k is k(k + 1) + k - k = k(k + 1), which is k² + k.

Method 4: The number of circles in Step k is k(k + 1), which equals k² + k.

Let us simplify each expression: Method 1: (k + 1)² - (k + 1) = k² + 2k + 1 - k - 1 = k² + k Method 2: k² + k Method 3: k(k + 1) = k² + k Method 4: k(k + 1) = k² + k

When carried out correctly, all methods lead to the same answer: k² + k, or equivalently k(k + 1). So the expression k(k + 1) gives the number of circles at Step k of this pattern.

So for Step 10, the number of circles would be 10 × (10 + 1) = 10 × 11 = 110.

Now, there is another pattern with square tiles. In Step 1, there is 1 tile. In Step 2, there are 4 tiles. In Step 3, there are 9 tiles. This appears to be square numbers: 1², 2², 3². But wait, that's not what the chapter says. Let me re-read. Actually, the chapter shows a different pattern. Let me focus on the expressions given.

From the pattern described in the "Figure it Out" section, we need to find the number of square tiles in each figure, in Step 4, in Step 10, and write an algebraic expression for Step n.

Based on the pattern, it seems like the number of tiles follows a square pattern: 1, 4, 9, 16,... which is n². So for Step n, the number of tiles would be n². For Step 4, it would be 16, and for Step 10, it would be 100.

Now, there is also a problem about finding the area of a shaded region. There are multiple methods given, and we need to verify that they all give the same answer.

Tadang's method: The total region is a square of side (m + n) with an area (m + n)². Subtracting the area of four rectangles from the total area will give the area of the interior shaded region. That is, (m + n)² - 4mn.

Yusuf's method: The shaded region is a square with side length (n - m). So its area is (n - m)².

We need to expand both expressions to check that they are equal. (m + n)² - 4mn = m² + 2mn + n² - 4mn = m² - 2mn + n² = (n - m)². Yes, they are equal!

Similarly, there is another problem about finding the area of a region with slanting lines. Three methods are given: Anusha's method, Vaishnavi's method, and Aditya's method. We need to verify that all three expressions are equivalent.

Anusha's method: Required area = Area(ABCD) - Area(EFGH) = x² - xy = x(x - y).

Vaishnavi's method: QS = y + x + y = x + 2y. Area of PQSR = x(x + 2y). Required area = x(x + 2y) - 3xy = x² + 2xy - 3xy = x² - xy = x(x - y).

Aditya's method: The required area is 2 times the area of JKLM. JK = (x - y)/2, KM = x. Area(JKML) = x((x - y)/2). Required area = 2 × x((x - y)/2) = x(x - y).

All three methods give the same answer: x(x - y). If x = 8 and y = 3, then the area is 8 × (8 - 3) = 8 × 5 = 40.

Now, let me summarize everything we have learned in this chapter.

In this chapter, we learned about the distributive property of multiplication over addition. We extended this property to find the product of two expressions each of which has two terms. The general form for the same is (a + b) × (c + d) = ac + ad + bc + bd.

We saw some special cases of this identity. First, the square of the sum of two numbers: (a + b)² = a² + 2ab + b². Second, the square of the difference of two numbers: (a - b)² = a² - 2ab + b². Third, the product of the sum and difference of two numbers: (a + b)(a - b) = a² - b².

We also learned how to use these identities to compute products and squares quickly. For example, we can compute 98 × 102 as (100 - 2)(100 + 2) = 10000 - 4 = 9996. We can compute 31² as (31 + 1)(31 - 1) + 1 = 32 × 30 + 1 = 961.

We considered different patterns and explored how to understand them using algebra. We saw that often there are multiple ways to solve a problem and arrive at the same correct answer. Finding different methods to approach and solve the same problem is a creative process, and it is important to explore different ways of thinking about mathematical problems.

We also learned about common mistakes to avoid when simplifying algebraic expressions, and how to identify and correct them.

This brings us to the end of our lesson. I hope you have understood the distributive property and its applications well. Remember, mathematics is not just about memorizing formulas but about understanding why they work and how they can be applied to solve problems. Keep practicing, and you will become more comfortable with these concepts.

Thank you for your attention, and goodbye till our next lesson!