NCERT Solutions Class 8 Maths Chapter 06 We Distribute, Yet Things Multiply

Question 1. By how much does the product increase if the first number (23) is increased by 1?
Answer: When the first number 23 is increased by 1, the multiplication becomes 24 × 27. Using the distributive property, (a + 1)b = ab + b, so the product goes up by 27.
In simple words: If you make the first number one bigger, the whole product gets bigger by exactly the second number.

Exam Tip: Remember that (a + 1)b = ab + b - this is the key distributive property that shows how adding 1 to one factor affects the product.

Question 2. What if the second number (27) is increased by 1?
Answer: If the second number 27 is increased by 1, the multiplication becomes 23 × 28. Applying a(b + 1) = ab + a, the product goes up by 23.
In simple words: When you make the second number one bigger, the product grows by the amount of the first number.

Exam Tip: This shows symmetry in multiplication - increasing either factor by 1 increases the product by the other factor.

Question 3. How about when both numbers are increased by 1? Do you see a pattern that could help generalise our observations to the product of any two numbers?
Answer: When both numbers are increased by 1, the multiplication becomes 24 × 28. Using the identity (a + 1)(b + 1) = ab + a + b + 1, the product goes up by 23 + 27 + 1 = 51. The pattern shows that when one number rises by 1, the product rises by the other number; when both rise by 1, the increase is a + b + 1.
In simple words: When you add 1 to both numbers, the product jumps by the first number plus the second number plus one more.

Exam Tip: The identity (a + 1)(b + 1) = ab + a + b + 1 is fundamental - it shows how two changes combine to affect a product.

Question 4. How do we expand (a + 1)(b + 1)?
Answer: To expand (a + 1)(b + 1), we distribute each term in the first bracket by each term in the second: (a + 1)(b + 1) = ab + a + b + 1.
In simple words: Multiply a by everything in the second bracket, then multiply 1 by everything in the second bracket, and add all the pieces together.

Exam Tip: Always use the distributive property systematically - multiply the first term by all terms in the second bracket, then the second term by all terms in the second bracket.

Question 5. What would we get if we had expanded (a + 1)(b + 1) by first taking (b + 1) as a single term?
Answer: We can treat (b + 1) as one single unit and apply the distributive property: (a + 1)(b + 1) = a(b + 1) + 1(b + 1). This method also leads to the same result.
In simple words: You can group part of the expression and treat it as a single thing - it gives the same answer either way.

Exam Tip: Different methods of expansion should always give the same final answer - if they don't, you made an arithmetic mistake.

Question 6. 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?
Answer: Let the two numbers be a and b with original product ab. If a is increased by 1 and b is decreased by 1, the new product becomes (a + 1)(b - 1) = ab - a + b - 1. The change in the product is b - a - 1. This means the product does not always change - it depends on the values of a and b. When b - a - 1 is positive, the product rises; when it equals zero, the product stays the same; when it is negative, the product falls.
In simple words: Adding 1 to one number and removing 1 from the other doesn't always keep the product the same - whether it gets bigger or smaller depends on which number was bigger to start with.

Exam Tip: The change formula b - a - 1 tells you exactly how and when the product changes - learn to apply this systematically for different number pairs.

Question 7. Will the product always increase? Find 3 examples where the product decreases.
Answer: No, the product will not always increase. From (a + 1)(b - 1) = ab + b - a - 1, the change in the product is b - a - 1. When this value is negative, the product shrinks. Below are three examples where the product shrinks.

Example 1: a = 10 and b = 8
Change = b - a - 1 = 8 - 10 - 1 = -3
Negative - product shrinks.
Original product = 10 × 8 = 80
New product = 11 × 7 = 77
It decreased.

Example 2: a = 15 and b = 12
Change = 12 - 15 - 1 = -4
Negative - product shrinks.
Original product = 15 × 12 = 180
New product = 16 × 11 = 176
It decreased.

Example 3: a = 7 and b = 5
Change = 5 - 7 - 1 = -3
Negative - product shrinks.
Original product = 7 × 5 = 35
New product = 8 × 4 = 32
It decreased.
In simple words: The product shrinks when you add 1 to a larger number and take away 1 from a smaller number.

Exam Tip: Always check whether b - a - 1 is positive, zero, or negative before concluding what happens to the product.

Question 8. What happens when a and b are negative integers?
Answer: When a and b are negative integers, the same rules apply as with positive numbers. The distributive property holds for all integers, so the identity (a + 1)(b - 1) = ab + b - a - 1 also works when a and b are negative. You simply put the negative values in place and follow the same steps. The product may go up or down depending on the value of b - a - 1, following the same pattern as with positive integers.
In simple words: The rules for how products change work the same way whether your numbers are positive or negative.

Exam Tip: Don't be scared of negative numbers - the algebraic identities still hold, and you can trust the formulas to work.

Question 9. By how much will the product of two numbers change if one of the numbers is increased by m and the other by n?
Answer: If one number is increased by m and the other by n, the new product becomes (a + m)(b + n). Expanding this, we get (a + m)(b + n) = ab + mb + an + mn. So the rise in the product is an + bm + mn. This formula tells us exactly how much the product shifts when the first number goes up by m and the second number goes up by n. This rule works for all integers.
In simple words: The product changes by taking each increase, multiplying it by the other number, plus the increases times each other.

Exam Tip: The formula an + bm + mn captures all possible product changes - memorize this key identity.

Question 10. Can you see how this identity can be used when one or both numbers are decreased?
Answer: Yes, we can use the same identity even when one or both numbers shrink. The identity is (a + m)(b + n) = ab + mb + an + mn. Here, m and n can be positive or negative. So if a number is decreased, we simply take m or n as a negative number. For example, if a goes up by 1 and b goes down by 1, then we take m = 1 and n = -1. Putting these into the identity: (a + 1)(b - 1) = ab + b - a - 1. So the identity works for both rises and falls, because decreases are written as adding a negative number.
In simple words: The identity works for both increases and decreases - just use negative numbers when something gets smaller.

Exam Tip: Treat decreases as negative increases - this unified approach makes all calculations consistent.

Question 11. Use Identity 1 to find how the product changes when (i) one number is decreased by 2 and the other increased by 3; (ii) both numbers are decreased, one by 3 and the other by 4.
Answer: Identity 1: (a + m)(b + n) = ab + mb + an + mn
The change in the product is mb + an + mn

(i) One number is decreased by 2 and the other is increased by 3
Decreased by 2 - m = -2
Increased by 3 - n = 3
Now replace m = -2 and n = 3 in the identity change formula:
Change = mb + an + mn
= (-2)b + 3a + (-2 × 3)
= -2b + 3a - 6
So the product changes by 3a - 2b - 6.

(ii) Both numbers are decreased, one by 3 and the other by 4
First number decreased by 3 - m = -3
Second number decreased by 4 - n = -4
Change = mb + an + mn
= (-3)b + (-4)a + (-3 × -4)
= -3b - 4a + 12
So the product changes by 12 - 4a - 3b.
In simple words: For decreases, use negative values for m and n; the formula stays exactly the same.

Exam Tip: Always identify whether each change is an increase (positive) or decrease (negative), then substitute into the formula carefully.

Question 12. Verify the answers by finding the products without converting the subtractions to additions.
Answer: Verification by direct products (without converting subtractions to additions)

We already found:
(i) Change = 3a - 2b - 6
(ii) Change = 12 - 4a - 3b

Now we verify each by multiplying directly.

(i) One number decreased by 2 and the other increased by 3
Original product: a × b = ab
New numbers: (a - 2) and (b + 3)
New product: (a - 2)(b + 3)
Now expand directly: (a - 2)(b + 3)
= a·b + a·3 - 2·b - 2·3
= ab + 3a - 2b - 6
Change in product = (new product - old product)
= (ab + 3a - 2b - 6) - ab
= 3a - 2b - 6
Hence Verified.

(ii) Both numbers decreased
First decreased by 3 - (a - 3)
Second decreased by 4 - (b - 4)
Original product: ab
New product: (a - 3)(b - 4)
Expand directly: (a - 3)(b - 4)
= a·b - 4a - 3b + 12
= ab - 4a - 3b + 12
Change in product = (new product - old product)
= (ab - 4a - 3b + 12) - ab
= 12 - 4a - 3b
Hence Verified.
In simple words: When you expand the products directly, you get the same changes - this proves the identity works.

Exam Tip: Always verify your identity results using direct multiplication - this shows you understand both methods and builds confidence in the formula.

Question 13. Expand (i) (a - u)(b + v), (ii) (a - u)(b - v).
Answer: (i) Using distributive property:
(a - u)(b + v)
= a·b + a·v - u·b - u·v
= ab + av - ub - uv

(ii) Using distributive property:
(a - u)(b - v)
= a·b - a·v - u·b + u·v
= ab - av - ub + uv
In simple words: When expanding, multiply each term in the first bracket by each term in the second bracket, keeping track of the plus and minus signs.

Exam Tip: Pay close attention to signs - a negative times a negative gives a positive, but a negative times a positive gives a negative.

Question 14. Example 1: Expand 3a/2 (a - b + 1/5).
Answer: 3a/2 (a - b + 1/5)
= (3a/2) × a - (3a/2) × b + (3a/2) × (1/5)
= 3/2 a² - 3/2 ab + 3/10 a
In simple words: Multiply the single term outside the bracket by each term inside, one at a time.

Exam Tip: When multiplying fractions by variables, handle the numbers and letters separately - it's easier to keep track.

Question 15. Can any two terms be added to get a single term?
Answer: No, we cannot add any two terms to make a single term. The terms we got were: 3/2 a², - 3/2 ab, and 3/10 a. All three terms have different letter parts: a², ab, and a. Since their letters are not the same, they are not like terms. Only like terms can be added.
In simple words: Two terms can only combine if they have exactly the same letters - a² and a are different, so they can't be added.

Exam Tip: Always check the letter part (variable part) before combining terms - the numbers can be different, but the variable parts must be identical.

Question 16. Example 2: Expand (a + b)(a + b).
Answer: First, distribute each term: (a + b)(a + b)
= a × a + b × a + a × b + b × b
= a² + ba + ab + b²
Now, ba and ab are like terms (both mean ab), so we add them: ba + ab = 2ab
So the final answer is a² + 2ab + b².
In simple words: When you expand and get repeated variable pairs, you can combine them into one term with a bigger number in front.

Exam Tip: Remember that ba and ab are the same - multiplication is commutative, so always look for these "hidden" like terms.

Question 17. Example 3: Expand (a + b)(a² + 2ab + b²).
Answer: Distributing (a + b) to each term inside the bracket: (a + b)(a² + 2ab + b²)
= (a + b)a² + (a + b)2ab + (a + b)b²
Now expanding each part:
- (a + b)a² = a·a² + b·a² = a³ + a²b
- (a + b)2ab = 2ab·a + 2ab·b = 2a²b + 2ab²
- (a + b)b² = a·b² + b·b² = ab² + b³
Now put all terms together: a³ + a²b + 2a²b + 2ab² + ab² + b³
Combine like terms:
a²b + 2a²b = 3a²b
ab² + 2ab² = 3ab²
So the final answer is a³ + 3a²b + 3ab² + b³.
In simple words: Distribute one group to each term in the other group, then combine any repeating patterns.

Exam Tip: For longer expansions, organize your work by distributing one piece at a time, then collect like terms in a second pass.

Question 1. Observe the multiplication grid below. Each number inside the grid is formed by multiplying two numbers. If the middle number of a 3 × 3 frame is given by the expression pq, as shown in the figure, write the expressions for the other numbers in the grid.
Answer:

Exam Tip: Understanding how a grid pattern works algebraically is important - each entry follows the same rule based on its position relative to the center.

Question 2. Expand the following products. (i) (3 + u)(v - 3) (ii) 2/3(15 + 6a) (iii) (10a + b)(10c + d) (iv) (3 - x)(x - 6) (v) (-5a + b)(c + d) (vi) (5 + z)(y + 9)
Answer:
(i) (3 + u)(v - 3)
= 3v - 9 + uv - 3u
= 3v + uv - 9 - 3u

(ii) 2/3(15 + 6a)
2/3 × 15 = 10
2/3 × 6a = 4a
So, 2/3(15 + 6a) = 10 + 4a

(iii) (10a + b)(10c + d)
(10a)(10c) = 100ac
(10a)(d) = 10ad
(b)(10c) = 10bc
(b)(d) = bd
So, (10a + b)(10c + d) = 100ac + 10ad + 10bc + bd

(iv) (3 - x)(x - 6)
(3)(x) = 3x
(3)(-6) = -18
(-x)(x) = -x²
(-x)(-6) = +6x
So, (3 - x)(x - 6) = 3x - 18 - x² + 6x
= -x² + 9x - 18

(v) (-5a + b)(c + d)
(-5a)(c) = -5ac
(-5a)(d) = -5ad
(b)(c) = bc
(b)(d) = bd
So, (-5a + b)(c + d) = -5ac - 5ad + bc + bd

(vi) (5 + z)(y + 9)
(5)(y) = 5y
(5)(9) = 45
(z)(y) = zy
(z)(9) = 9z
So, (5 + z)(y + 9) = 5y + 45 + zy + 9z
In simple words: Multiply each term in the first group by each term in the second group, then collect and organize all the results.

Exam Tip: Write out each multiplication piece separately before combining - this prevents sign errors and makes your work easy to check.

Question 3. Find 3 examples where the product of two numbers remains unchanged when one of them is increased by 2 and the other is decreased by 4.
Answer: We want the product to remain the same when:
- one number is increased by 2, and
- the other is decreased by 4.

So if the original product is a × b
The new product becomes: (a + 2)(b - 4)
For the product to stay unchanged: (a + 2)(b - 4) = ab
\( \implies \) ab - 4a + 2b - 8 = ab
\( \implies \) -4a + 2b - 8 = 0
\( \implies \) 2b - 4a - 8 = 0
\( \implies \) b - 2a - 4 = 0
So, b = 2a + 4

This means any pair of numbers (a, b) that satisfy b = 2a + 4 will keep the product unchanged.

Example 1
Let a = 1
Then b = 2(1) + 4 = 6
Original product: 1 × 6 = 6
New product: (1 + 2)(6 - 4) = 3 × 2 = 6
Product unchanged.

Example 2
Let a = 2
Then b = 2(2) + 4 = 8
Original product: 2 × 8 = 16
New product: (2 + 2)(8 - 4) = 4 × 4 = 16
Product unchanged.

Example 3
Let a = 3
Then b = 2(3) + 4 = 10
Original product: 3 × 10 = 30
New product: (3 + 2)(10 - 4) = 5 × 6 = 30
Product unchanged.
In simple words: You can keep a product the same by increasing one number and decreasing the other, but only if the second number starts off bigger in a special way.

Exam Tip: Set up the equation (a + 2)(b - 4) = ab and solve for the relationship between a and b - this gives you all possible pairs that work.

Question 4. Expand (i) (a + ab - 3b²)(4 + b), and (ii) (4y + 7)(y + 11z - 3).
Answer: (i) Using distributive property
(a + ab - 3b²)(4 + b)
= (a + ab - 3b²) × 4 + (a + ab - 3b²) × b
= (4a + 4ab - 12b²) + (ab + ab² - 3b³)
= 4a + 5ab + ab² - 12b² - 3b³

(ii) Using distributive property
(4y + 7)(y + 11z - 3)
= 4y(y + 11z - 3) + 7(y + 11z - 3)
= (4y² + 44yz - 12y) + (7y + 77z - 21)
= 4y² + 44yz - 5y + 77z - 21
In simple words: Multiply the whole first group by each term in the second group, then add all the results together.

Exam Tip: When expanding, break the work into steps - distribute to one term at a time, then combine like terms at the end.

Question 5. Expand (i) (a - b)(a + b), (ii) (a - b)(a² + ab + b²) and (iii) (a - b)(a³ + a²b + ab² + b³). Do you see a pattern? What would be the next identity in the pattern that you see? Can you check it by expanding?
Answer: (i) Using distributive property:
(a - b)(a + b)
= a² - b²

(ii) Using distributive property:
(a - b)(a² + ab + b²)
= a³ + a²b + ab² - (a²b + ab² + b³)
= a³ - b³

(iii) Using distributive property:
(a - b)(a³ + a²b + ab² + b³)
= a⁴ + a³b + a²b² + ab³ - (a³b + a²b² + ab³ + b⁴)
= a⁴ - b⁴

Yes there is a pattern. Every time we multiply:
(a - b) × (sum of all mixed powers of a and b)
we get: aⁿ - bⁿ

For n = 2 - a² - b²
For n = 3 - a³ - b³
For n = 4 - a⁴ - b⁴

The next identity will be:
(a - b)(a⁴ + a³b + a²b² + ab³ + b⁴)
= a⁵ - b⁵

Checking by expanding:
(a - b)(a⁴ + a³b + a²b² + ab³ + b⁴)
= a⁵ + a⁴b + a³b² + a²b³ + ab⁴ - (a⁴b + a³b² + a²b³ + ab⁴ + b⁵)
= a⁵ - b⁵
The pattern is correct.
In simple words: The pattern shows that (a - b) times a staircase of powers always gives a⁵ minus b⁵, a⁶ minus b⁶, and so on.

Exam Tip: Spotting and proving patterns is a key algebra skill - always verify your pattern by expanding the next case before claiming it's true.

Question 1. Use the following multiplications to find the product of a number with 11 in a single step. (a) 3874 × 11 (b) 5678 × 11
Answer: (a) 3874 × 11
Writing the number with space between digits: 3 8 7 4
Now insert sums:
- Between 3 and 8 - 3 + 8 = 11
- Between 8 and 7 - 8 + 7 = 15
- Between 7 and 4 - 7 + 4 = 11
Now write step by step with carries:
Start with last digit - 4
Next - 7 + 4 = 11 - write 1, carry 1
Next - 8 + 7 + carry (1) = 16 - write 6, carry 1
Next - 3 + 8 + carry (1) = 12 - write 2, carry 1
First digit + carry - 3 + carry (1) = 4
Final number: 42614
So, 3874 × 11 = 42614

(b) 5678 × 11
Digits: 5 6 7 8
Sums:
- 5 + 6 = 11
- 6 + 7 = 13
- 7 + 8 = 15
Now apply carries:
Last digit - 8
7 + 8 = 15 - write 5, carry 1
6 + 7 + 1 = 14 - write 4, carry 1
5 + 6 + 1 = 12 - write 2, carry 1
First digit + carry - 5 + 1 = 6
Final answer: 62458
So, 5678 × 11 = 62458
In simple words: To multiply by 11, add up each pair of side-by-side digits and write each sum in between, handling carries when a sum goes over 9.

Exam Tip: The carry step is critical - when a sum reaches 10 or more, you must write down only the ones digit and carry the rest forward.

Question 2. Describe a general rule to multiply a number (of any number of digits) by 11 and write the product in one line. Evaluate (i) 94 × 11, (ii) 495 × 11, (iii) 3279 × 11, (iv) 4791256 × 11.
Answer: To multiply any number by 11 in one line:
- Write the first digit.
- Add each pair of neighbouring digits and write the sums in order.
- Write the last digit.
- If any sum is 10 or more, carry over the extra digit.

(i) 94 × 11
Digits: 9 4
Sum: 9 + 4 = 13
Write with carries:
- Last digit - 4
- Middle - 13 - write 3, carry 1
- First digit + carry - 9 + 1 = 10
So, 94 × 11 = 1034

(ii) 495 × 11
Digits: 4 9 5
Sums: 4 + 9 = 13, 9 + 5 = 14
Write with carries:
- Last digit - 5
- 9 + 5 = 14 - write 4, carry 1
- 4 + 9 + 1 = 14 - write 4, carry 1
- First digit + carry - 4 + 1 = 5
So, 495 × 11 = 5445

(iii) 3279 × 11
Digits: 3 2 7 9
Sums:
3 + 2 = 5
2 + 7 = 9
7 + 9 = 16
Now write:
- Last digit - 9
- 7 + 9 = 16 - write 6, carry 1
- 2 + 7 + 1 = 10 - write 0, carry 1
- 3 + 2 + 1 = 6
- First digit - 3
So, 3279 × 11 = 36069

(iv) 4791256 × 11
Digits: 4 7 9 1 2 5 6
Sums:
4 + 7 = 11
7 + 9 = 16
9 + 1 = 10
1 + 2 = 3
2 + 5 = 7
5 + 6 = 11
Now place digits and carry:
- Last digit - 6
- 5 + 6 = 11 - write 1, carry 1
- 2 + 5 + 1 = 8
- 1 + 2 = 3
- 9 + 1 = 10 - write 0, carry 1
- 7 + 9 + 1 = 17 - write 7, carry 1
- 4 + 7 + 1 = 12 - write 2, carry 1
- First digit + carry - 4 + 1 = 5
So, 4791256 × 11 = 52703716
In simple words: Add neighbours and write sums with carries - that's the whole trick for multiplying by 11.

Exam Tip: Write down the digit sums clearly as a first step, then apply carries in a second pass - this keeps errors down and makes checking easier.

Question 3. Can we come up with a similar rule for multiplying a number by 101? Multiply 3874 by 101. Use this to multiply 3874 × 101 in one line.
Answer: Yes, we can make a similar rule for multiplying a number by 101.
It is given that: number × 101 = number × (100 + 1)
So, to multiply any number by 101:
Write the number with two zeros added to the right.
Write the same number below it.
Add the two numbers.
This gives the product in one step.

Multiplication of 3874 by 101 = 3874 × 101 = 391274

Multiplication of 3874 × 101 in one line:
3874 × 101 = 387400 + 3874 = 391274
This method works for any number.
In simple words: To multiply by 101, write the number, add two zeros after it, write the number again below, and add the two lines together.

Exam Tip: This method works because 101 = 100 + 1, so you're really just writing the number shifted left by 100 places, plus the original number.

Question 4. What could be a general rule to multiply a number by 101 and write the product in one line? Extend this rule for multiplication by 1001, 10001, …
Answer: A general rule to multiply any number by 101 is:
- Write the number.
- Write the same number again, but shifted two places to the right (like adding two zeros).
- Add the two numbers.
This gives the product in one line.

Example:
3874 × 101 = 387400 + 3874 = 391274

This idea can be extended: For × 1001:
- Write the number.
- Write the same number again, shifted three places to the right.
- Add them.

For × 10001:
- Write the number.
- Write the same number again, shifted four places to the right.
- Add them.

So for multiplication by 101, 1001, 10001, …
just write the number twice and the second copy is shifted according to the number of zeros.
In simple words: The number of zeros tells you how many places to shift - 101 has one zero so shift by 2, 1001 has two zeros so shift by 3, and so on.

Exam Tip: This pattern works because 101 = 100 + 1, 1001 = 1000 + 1, and so on - the "+1" creates the original number, and the "100" or "1000" creates the shifted copy.

Question 5. Use this to find (i) 89 × 101, (ii) 949 × 101, (iii) 265831 × 1001, (iv) 1111 × 1001, (v) 9734 × 99 and (vi) 23478 × 999.
Answer: (i) 89 × 101
Using the rule: number × 101 = number with two zeros + number
So, 89 × 101
= 8900 + 89
= 8989

(ii) 949 × 101
= 94900 + 949
= 95849

(iii) 265831 × 1001
Using the Rule: number × 1001 = number with three zeros + number
265831 × 1001
= 265831000 + 265831
= 266096831

(iv) 1111 × 1001
= 1111000 + 1111
= 1112111

(v) 9734 × 99
We know that: 99 = 100 - 1
So use distributive property:
9734 × 99
= 9734 × (100 - 1)
= 973400 - 9734
= 963666

(vi) 23478 × 999
Putting 999 = 1000 - 1, we have
23478 × 999
= 23478 × (1000 - 1)
= 23478000 - 23478
= 23454522
In simple words: For 101 and 1001, write the number twice with a shift; for 99 and 999, write the number shifted minus the original number.

Exam Tip: Choose the method based on whether the multiplier is one more (like 101) or one less (like 99) than a power of 10.

Question 6. The area of a square of sidelength 60 units is 3600 sq. units (60²) and that of a square of sidelength 5 units is 25 sq. units (5²). Can we use this to find the area of a square of sidelength 65 units?
Answer: Yes, we can use the areas of 60² and 5² to find the area of a square of side 65. Since 65 = 60 + 5, we can split the big square into: one square of side 60, one square of side 5, and two rectangles of sides 60 and 5. So the area is: (60 + 5)² = 60² + 5² + 2 × 60 × 5 = 3600 + 25 + 600 = 4225 sq. units. So, the area of the square with sidelength 65 is 4225 square units.
In simple words: You can build the large square using the two smaller squares plus two rectangles - that's where the (a + b)² formula comes from geometrically.

Exam Tip: Drawing a picture of how a large square breaks into smaller pieces helps you understand why (a + b)² = a² + b² + 2ab - it's not just a formula, it's a real geometric fact.

Question 7. What if we write 65² as (30 + 35)² or (52 + 13)²?
Answer: If we write 65² as (30 + 35)² or (52 + 13)², we will still get the same area because all these pairs add up to 65. Using the formula: (a + b)² = a² + b² + 2ab

For (30 + 35)²:
(30 + 35)²
= 30² + 35² + 2 × 30 × 35
= 900 + 1225 + 2100
= 4225

For (52 + 13)²:
(52 + 13)²
= 52² + 13² + 2 × 52 × 13
= 2704 + 169 + 1352
= 4225

So, no matter how we split 65, 65² = 4225 sq. units.
In simple words: No matter which two numbers you pick that add to 65, squaring their sum always gives 4225 - the answer is fixed, only the way you break it up changes.

Exam Tip: This shows the power of the algebraic identity - once you know (a + b)², you know that result for ANY split of any number into two parts.

Question 8. If a and b are any two integers, is (a + b)² always greater than a² + b²? If not, when is it greater?
Answer: (a + b)² is not always greater than a² + b². We know that: (a + b)² = a² + b² + 2ab. So, the difference between them is: (a + b)² - (a² + b²) = 2ab. This means: if 2ab > 0, then (a + b)² is greater; if 2ab = 0, then they are equal; if 2ab < 0, then (a + b)² is smaller. Now, when is 2ab > 0? When a and b have the same sign (both positive or both negative). So, (a + b)² is greater than a² + b² when a and b are either both positive or both negative. If one is positive and the other is negative, the product ab becomes negative and the result is not greater.
In simple words: (a + b)² is bigger than a² + b² only when both numbers have the same sign - if they have opposite signs, (a + b)² might be smaller.

Exam Tip: The key insight is looking at the term 2ab - its sign determines whether (a + b)² is larger or smaller, and the signs of a and b control the sign of 2ab.

Question 9. Use Identity 1A to find the values of 104², 37². (Hint: Decompose 104 and 37 into sums or differences of numbers whose squares are easy to compute.)
Answer: Identity 1A: (a + b)² = a² + b² + 2ab
We must break the numbers into easy parts.

104²
104 can be written as: 104 = 100 + 4
Now apply Identity 1A: (100 + 4)²
= 100² + 4² + 2 × 100 × 4
= 10000 + 16 + 800
= 10816

37²
37 can be written as: 37 = 40 - 3
Using Identity 1B: (a - b)² = a² + b² - 2ab
(40 - 3)²
= 40² + 3² - 2 × 40 × 3
= 1600 + 9 - 240
= 1369
In simple words: Pick a nearby round number and show your number as that round number plus or minus a small number - then use the identity to square it quickly.

Exam Tip: Break big numbers smartly - 104 breaks as 100 + 4 (good), and 37 breaks as 40 - 3 (good) - find the nearest "easy" square to minimize your arithmetic.

Question. Use Identity 1A to write the expressions for the following.
(i) \( (m + 3)^2 \)
(ii) \( (6 + p)^2 \)
Answer:
(i) Using Identity 1A: \( (m + 3)^2 \)
\( = m^2 + 3^2 + 2 \cdot m \cdot 3 \)
\( = m^2 + 9 + 6m \)
\( = m^2 + 6m + 9 \)
(ii) Using Identity 1A: \( (6 + p)^2 \)
\( = 6^2 + p^2 + 2 \cdot 6 \cdot p \)
\( = 36 + p^2 + 12p \)
\( = p^2 + 12p + 36 \)
In simple words: To expand a sum squared, you add the first term squared, the second term squared, and twice the product of both terms.

Exam Tip: Always remember the identity \( (a + b)^2 = a^2 + 2ab + b^2 \) - forgetting the middle term is a common mistake.

Question. Expand \( (6x + 5)^2 \).
Answer: We use Identity 1A: \( (a + b)^2 = a^2 + b^2 + 2ab \)
Here, \( a = 6x \) and \( b = 5 \)
So, \( (6x + 5)^2 \)
\( = (6x)^2 + 5^2 + 2 \cdot (6x) \cdot 5 \)
\( = 36x^2 + 25 + 60x \)
\( = 36x^2 + 60x + 25 \)
In simple words: Square the first part (which gives 36x²), square the second part (which gives 25), then add twice their product (which gives 60x).

Exam Tip: Write down each part of the identity separately before combining terms - this reduces careless errors with coefficients and powers.

Question. Expand \( (3j + 2k)^2 \) using both the identity and by applying the distributive property.
Answer: Using Identity 1A: \( (a + b)^2 = a^2 + b^2 + 2ab \)
Here, \( a = 3j \) and \( b = 2k \)
So, \( (3j + 2k)^2 \)
\( = (3j)^2 + (2k)^2 + 2 \cdot (3j)(2k) \)
\( = 9j^2 + 4k^2 + 12jk \)
\( = 9j^2 + 12jk + 4k^2 \)

Using Distributive Property:
\( (3j + 2k)(3j + 2k) \)
\( = 3j(3j + 2k) + 2k(3j + 2k) \)
\( = (9j^2 + 6jk) + (6jk + 4k^2) \)
\( = 9j^2 + 12jk + 4k^2 \)
In simple words: The identity method uses a formula, while the distributive property method multiplies each part of the first bracket by each part of the second bracket - both give the same answer.

Exam Tip: Showing both methods demonstrates stronger understanding and ensures your answer is correct if one approach confirms the other.

Question. Can we use 60² (= 3600) and 5² (= 25) to find the value of (60 - 5)² or 55²?
Answer: Yes, we can use 60² and 5² to find the value of (60 - 5)² or 55².
We write: \( 55 = 60 - 5 \)
Now use the identity for \( (a - b)^2 \): \( (a - b)^2 = a^2 + b^2 - 2ab \)
So, \( (60 - 5)^2 \)
\( = 60^2 + 5^2 - 2 \times 60 \times 5 \)
\( = 3600 + 25 - 600 \)
\( = 3025 \)
So, \( 55^2 = 3025 \)
In simple words: By breaking 55 into 60 minus 5, we can use the squares and the product of 60 and 5 to find our answer quickly, without multiplying 55 × 55 directly.

Exam Tip: Look for numbers close to round figures (like 50, 100) and break them as additions or subtractions to use the identity and mental calculation.

Question. We can also use the expansion of \( (a + b)^2 \) to find the expansion of \( (a - b)^2 \). Think how.
Answer: Yes, we can use the expansion of \( (a + b)^2 \) to derive the expansion of \( (a - b)^2 \).
The idea is: \( (a - b)^2 \) can be written as \( (a + (-b))^2 \)
Now apply the formula for \( (a + b)^2 \): \( (a + b)^2 = a^2 + b^2 + 2ab \)
Here, \( b \) is replaced by \( (-b) \):
\( (a + (-b))^2 \)
\( = a^2 + (-b)^2 + 2 \cdot a \cdot (-b) \)
\( = a^2 + b^2 - 2ab \)
So, \( (a - b)^2 = a^2 + b^2 - 2ab \)
This is the expansion of \( (a - b)^2 \), obtained directly from \( (a + b)^2 \).
In simple words: By treating the minus sign as adding a negative number, you can rewrite the subtraction formula using the addition formula you already know.

Exam Tip: Understanding how one identity follows from another shows deeper mathematical thinking and helps you remember formulas by their relationships.

Question. Find the general expansion of \( (a - b)^2 \) using geometry, as we did for 55².
Answer: To get the general expansion of \( (a - b)^2 \) using geometry, we imagine a big square of side \( a \) and remove two rectangles of side \( a \) and \( b \), just like we did for 60 and 5.
Start with a square of side \( a \): Area = \( a^2 \)
Inside it, imagine a smaller square of side \( b \) in one corner.
When we remove the two rectangles of size \( a \times b \), we have removed the small \( b \times b \) square twice. So we add it back once.
So the area becomes: \( a^2 - ab - ab + b^2 = a^2 - 2ab + b^2 \)
Final geometric expansion: \( (a - b)^2 = a^2 - 2ab + b^2 \)
This is the same idea used for 55².
In simple words: When you remove two rectangles from a large square, you accidentally take away the corner piece twice, so you have to add it back once.

Exam Tip: Drawing and labeling the geometric figure helps you see why the formula works and makes it easier to remember.

Question. Use the identity \( (a - b)^2 \) to find the values of (a) 99² and (b) 58².
Answer: Identity: \( (a - b)^2 = a^2 + b^2 - 2ab \)

(a) 99²
99 can be written as: \( 99 = 100 - 1 \)
Now applying the identity: \( (100 - 1)^2 \)
\( = 100^2 + 1^2 - 2 \times 100 \times 1 \)
\( = 10000 + 1 - 200 \)
\( = 9801 \)
So, \( 99^2 = 9801 \)

(b) 58²
58 can be written as: \( 58 = 60 - 2 \)
Now applying the identity: \( (60 - 2)^2 \)
\( = 60^2 + 2^2 - 2 \times 60 \times 2 \)
\( = 3600 + 4 - 240 \)
\( = 3364 \)
So, \( 58^2 = 3364 \)
In simple words: Write the number as a round number plus or minus a small number, then use the formula to calculate quickly.

Exam Tip: Pick round numbers (100, 60, 50) that are closest to your target number - the closer they are, the easier the mental arithmetic becomes.

Question. Expand the following using both Identity 1B and by applying the distributive property: (i) \( (b - 6)^2 \) (ii) \( (-2a + 3)^2 \) (iii) \( (7y - \frac{3}{4}z)^2 \)
Answer: Identity 1B: \( (a - b)^2 = a^2 - 2ab + b^2 \)

(i) \( (b - 6)^2 \)
Using Identity 1B where \( a = b, b = 6 \):
\( (b - 6)^2 = b^2 - 2 \cdot b \cdot 6 + 6^2 = b^2 - 12b + 36 \)
Using Distributive Property:
\( (b - 6)(b - 6) = b \cdot b - 6 \cdot b - 6 \cdot b + 36 = b^2 - 12b + 36 \)

(ii) \( (-2a + 3)^2 \)
Rewrite it as: \( (3 - 2a)^2 \)
Using Identity 1B where \( a = 3, b = 2a \):
\( (3 - 2a)^2 = 3^2 - 2 \cdot 3 \cdot (2a) + (2a)^2 = 9 - 12a + 4a^2 \)
Using Distributive Property:
\( (3 - 2a)(3 - 2a) = 3 \cdot 3 - 6a - 6a + 4a^2 = 9 - 12a + 4a^2 \)

(iii) \( (7y - \frac{3}{4}z)^2 \)
Let \( a = 7y, b = \frac{3}{4}z \)
Using Identity 1B:
\( (7y - \frac{3}{4}z)^2 = (7y)^2 - 2 \cdot (7y)(\frac{3}{4}z) + (\frac{3}{4}z)^2 = 49y^2 - \frac{21}{2}yz + \frac{9}{16}z^2 \)
Using Distributive Property:
\( (7y - \frac{3}{4}z)(7y - \frac{3}{4}z) = 49y^2 - \frac{21}{2}yz - \frac{21}{2}yz + \frac{9}{16}z^2 = 49y^2 - \frac{21}{2}yz + \frac{9}{16}z^2 \)
In simple words: Both methods (using the formula and multiplying directly) produce the same result - choose whichever feels easier for you.

Exam Tip: When variables or fractions are involved, be extra careful with signs and keep track of all terms - the second method helps you verify.

Question. Take a pair of natural numbers. Calculate the sum of their squares. Can you write twice this sum as a sum of two squares?
Answer: Let any pair of natural numbers be \( a \) and \( b \).
First calculate the sum of their squares: \( a^2 + b^2 \)
Now take twice this sum: \( 2(a^2 + b^2) \)
Now, \( (a + b)^2 + (a - b)^2 \)
\( = (a^2 + 2ab + b^2) + (a^2 - 2ab + b^2) \)
\( = 2a^2 + 2b^2 \)
\( = 2(a^2 + b^2) \)
So, \( 2(a^2 + b^2) = (a + b)^2 + (a - b)^2 \)
This means twice the sum of squares can always be written as the sum of two squares.

Example: Take the numbers 5 and 6.
Sum of their squares: \( 5^2 + 6^2 = 25 + 36 = 61 \)
Twice this sum: \( 2 \times 61 = 122 \)
Now write it as sum of two squares:
\( (5 + 6)^2 + (5 - 6)^2 \)
\( = 11^2 + (-1)^2 \)
\( = 121 + 1 \)
\( = 122 \)
In simple words: If you double the sum of any two numbers squared, you can split that result into the sum of two different squared numbers.

Exam Tip: Always verify the pattern with a concrete example before accepting it as true - this builds confidence in your understanding.

Question. Do the identities below help in explaining the observed pattern? \( (a + b)^2 = a^2 + 2ab + b^2 \) and \( (a - b)^2 = a^2 - 2ab + b^2 \)
Answer: Yes, these identities help explain the pattern.
We have:
\( (a + b)^2 = a^2 + 2ab + b^2 \)
\( (a - b)^2 = a^2 - 2ab + b^2 \)
Now add both identities: \( (a + b)^2 + (a - b)^2 \)
\( = (a^2 + 2ab + b^2) + (a^2 - 2ab + b^2) \)
The terms \( +2ab \) and \( -2ab \) cancel out. So we get:
\( = a^2 + a^2 + b^2 + b^2 \)
\( = 2a^2 + 2b^2 \)
\( = 2(a^2 + b^2) \)
This shows exactly why: \( 2(a^2 + b^2) = (a + b)^2 + (a - b)^2 \)
So the pattern is explained by these identities.
In simple words: When you add the two squared expressions together, the middle terms disappear because they are opposites, leaving you with double the original sum.

Exam Tip: Highlight how the \( +2ab \) and \( -2ab \) cancel - this is the key reason the formula works.

Question. 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
Answer: Looking at the given pattern:
\( 9 \times 9 - 1 \times 1 = 10 \times 8 \)
\( 8 \times 8 - 6 \times 6 = 14 \times 2 \)
\( 7 \times 7 - 2 \times 2 = 9 \times 5 \)
\( 10 \times 10 - 4 \times 4 = 14 \times 6 \)
On the left side, we always have: \( a \times a - b \times b = a^2 - b^2 \)
On the right side, we have: \( (a + b)(a - b) \)
So the pattern is: \( a^2 - b^2 = (a + b)(a - b) \)
Now we check this using algebra (distributive property):
\( (a + b)(a - b) \)
\( = a \cdot a - a \cdot b + b \cdot a - b \cdot b \)
\( = a^2 - ab + ab - b^2 \)
\( = a^2 - b^2 \)
So the two sides are equal.
Therefore, the pattern \( a^2 - b^2 = (a + b)(a - b) \) always holds for any numbers \( a \) and \( b \).
In simple words: The difference of two squared numbers equals the product of their sum and their difference - this works for every pair of numbers you pick.

Exam Tip: Always verify the identity algebraically after spotting the pattern - this proves it holds for all values, not just the examples shown.

Question. Use Identity 1C to calculate 98 × 102, and 45 × 55.
Answer: Identity 1C: \( (a + b)(a - b) = a^2 - b^2 \)

For 98 × 102:
98 and 102 are equally spaced around 100.
So, \( 98 = 100 - 2 \)
\( 102 = 100 + 2 \)
Now using Identity 1C:
\( (100 - 2)(100 + 2) \)
\( = 100^2 - 2^2 \)
\( = 10000 - 4 \)
\( = 9996 \)
So, \( 98 \times 102 = 9996 \)

For 45 × 55:
Here the midpoint is 50:
\( 45 = 50 - 5 \)
\( 55 = 50 + 5 \)
Now use Identity 1C:
\( (50 - 5)(50 + 5) \)
\( = 50^2 - 5^2 \)
\( = 2500 - 25 \)
\( = 2475 \)
So, \( 45 \times 55 = 2475 \)
In simple words: Find a round number that sits in the middle of your two numbers, break them as add/subtract from that middle number, then use squaring to finish the calculation.

Exam Tip: This identity turns a tricky multiplication into simple subtraction of squares - use it whenever you see two numbers equally spaced from a round value.

Question. Show that \( (a + b) \times (a - b) = a^2 - b^2 \) geometrically.
Answer: To show geometrically that \( (a + b)(a - b) = a^2 - b^2 \), we use a square picture.
Draw a big square of side \( a \). Its area is \( a^2 \).
Inside it, remove a smaller square of side \( b \) from one corner. The area removed is \( b^2 \).
The remaining shape looks like a rectangle. Its length is \( a + b \) and its width is \( a - b \).
When we remove the \( b \times b \) square, one side becomes shorter by \( b \) and the other side becomes longer by \( b \).
So the area of the remaining rectangle is: \( (a + b)(a - b) \)
But this area is also equal to: \( a^2 - b^2 \)
Since both represent the same area, we get \( (a + b)(a - b) = a^2 - b^2 \)
This shows the identity geometrically.
In simple words: Start with a large square, cut out a small square from the corner, and the leftover piece is a rectangle whose length and width multiply to give the same answer as big square minus small square.

Exam Tip: Draw and label the figure carefully - the visual proof often helps you understand why the algebraic identity is true.

Question. Why is this identity true?
Answer: The identity \( (a + b)(a - b) = a^2 - b^2 \) is true because both sides represent the same area in two different ways.
Think of a big square of side \( a \). Its area is \( a^2 \).
Inside it, take out a small square of side \( b \). The area removed is \( b^2 \).
So the remaining area is \( a^2 - b^2 \), but this remaining shape can also be seen as a rectangle.
Its length becomes \( (a + b) \) and its width becomes \( (a - b) \).
So the area is also \( (a + b)(a - b) \).
Since both expressions describe the same region, their areas must be equal.
In simple words: The same leftover piece can be measured two ways - by subtracting areas, or by multiplying the rectangle's length and width - both ways give the same answer.

Exam Tip: Remember that geometric proofs show WHY a formula is true, while algebraic proofs show that it ALWAYS works for any numbers.

Question 1. Which is greater: \( (a - b)^2 \) or \( (b - a)^2 \)? Justify your answer.
Answer: \( (a - b)^2 \) and \( (b - a)^2 \) are always equal.
This is because when we square a number, the sign disappears.
For example, \( 5^2 = 25 \) and \( (-5)^2 = 25 \).
Here, \( b - a = -(a - b) \)
So, \( (b - a)^2 = [-(a - b)]^2 \)
\( = (a - b)^2 \)
Both are the same.
Therefore, neither is greater. \( (a - b)^2 = (b - a)^2 \) for all values of \( a \) and \( b \).
In simple words: When you square any number, whether it is positive or negative, you always get a positive result - so swapping the order in a subtraction and squaring gives you the same answer.

Exam Tip: This property is useful when simplifying expressions - if you see \( (x - y)^2 \), you know it equals \( (y - x)^2 \) without any further work.

Question 2. Express 100 as the difference of two squares.
Answer: We have to write 100 as a difference of two squares.
\( 100 = a^2 - b^2 \)
Now find two numbers whose product is 100 using: \( a^2 - b^2 = (a + b)(a - b) \)
Taking the pair: \( 25 \times 4 = 100 \)
So, \( a + b = 25 \) and \( a - b = 4 \)
Add the equations: \( 2a = 29 \Rightarrow a = 14.5 \) and \( b = 10.5 \)
Now: \( a^2 - b^2 = 14.5^2 - 10.5^2 \)
\( = (210.25 - 110.25) \)
\( = 100 \)
Therefore, \( 100 = 14.5^2 - 10.5^2 \)
In simple words: Find two factors of 100 that fit the pattern (a + b)(a - b), solve for a and b, then square them to verify.

Exam Tip: Remember that a and b do not have to be whole numbers - decimals and fractions work just as well in this identity.

Question 3. Find 406², 72², 145², 1097² and 124² using the identities you have learnt so far.
Answer:
For 406²:
\( 406 = 400 + 6 \)
\( \Rightarrow (400 + 6)^2 = 400^2 + 6^2 + 2 \cdot 400 \cdot 6 = 160000 + 36 + 4800 = 164836 \)
So, \( 406^2 = 164836 \)

For 72²:
\( 72 = 70 + 2 \)
\( \Rightarrow (70 + 2)^2 = 70^2 + 2^2 + 2 \cdot 70 \cdot 2 = 4900 + 4 + 280 = 5184 \)
So, \( 72^2 = 5184 \)

For 145²:
\( 145 = 150 - 5 \)
\( \Rightarrow (150 - 5)^2 = 150^2 + 5^2 - 2 \cdot 150 \cdot 5 = 22500 + 25 - 1500 = 21025 \)
So, \( 145^2 = 21025 \)

For 1097²:
\( 1097 = 1100 - 3 \)
\( \Rightarrow (1100 - 3)^2 = 1100^2 + 3^2 - 2 \cdot 1100 \cdot 3 = 1210000 + 9 - 6600 = 1203409 \)
So, \( 1097^2 = 1203409 \)

For 124²:
\( 124 = 120 + 4 \)
\( \Rightarrow (120 + 4)^2 = 120^2 + 4^2 + 2 \cdot 120 \cdot 4 = 14400 + 16 + 960 = 15376 \)
So, \( 124^2 = 15376 \)
In simple words: For each number, write it as a round number plus or minus a small number, then apply the appropriate squared formula to calculate quickly.

Exam Tip: Always choose the nearest round number (multiples of 10 or 100) - this keeps the arithmetic simple and reduces errors.

Question 4. Do Patterns 1 and 2 hold only for counting numbers? Do they hold for negative integers as well? What about fractions? Justify your answer.
Answer: Patterns 1 and 2 come from the identities:
\( (a + b)^2 = a^2 + b^2 + 2ab \)
\( (a - b)^2 = a^2 + b^2 - 2ab \)
\( (a + b)(a - b) = a^2 - b^2 \)
These identities do not depend on \( a \) and \( b \) being counting numbers only.
They work for:
- counting numbers (1, 2, 3, ...)
- negative integers
- fractions
- all real numbers
Because the distributive property works for all numbers.

Taking Pattern 2: \( a^2 - b^2 = (a + b)(a - b) \)
Taking \( a = -5, b = 2 \):
Left side: \( (-5)^2 - 2^2 = 25 - 4 = 21 \)
Right side: \( (-5 + 2)(-5 - 2) = (-3)(-7) = 21 \)
Both sides match.
So the pattern works for negative integers too.

Now, taking \( a = \frac{1}{2} \) and \( b = \frac{1}{3} \):
Left side: \( (\frac{1}{2})^2 - (\frac{1}{3})^2 = \frac{1}{4} - \frac{1}{9} = \frac{5}{36} \)
Right side: \( (\frac{1}{2} + \frac{1}{3})(\frac{1}{2} - \frac{1}{3}) = (\frac{5}{6})(\frac{1}{6}) = \frac{5}{36} \)
Again both sides match.
Therefore, Patterns 1 and 2 hold for all numbers, not just counting numbers, because the algebraic identities behind them are true for every integer, fraction and real number.
In simple words: The rules work for any kind of number you can think of - positive whole numbers, negative numbers, fractions, and decimals - because they are based on how multiplication and addition work.

Exam Tip: Always back up your claims with numerical examples showing negative numbers and fractions work too - this demonstrates complete understanding.

Mistake Identification and Correction

Question. We have expanded and simplified some algebraic expressions below to their simplest forms. (i) Check each of the simplifications and see if there is a mistake. (ii) If there is a mistake, try to explain what could have gone wrong. (iii) Then write the correct expression. [12 expressions given]
Answer:

Expression 1: \( -3p (-5p + 2q) \)
Given: \( -3p + 5p - 2q = p - 2q \)
Mistake: The student did not multiply \( p \) with \( p \) and \( q \) - they subtracted instead of multiplying.
Correct: \( -3p(-5p + 2q) = 15p^2 - 6pq \)

Expression 2: \( 2(x - 1) + 3(x + 4) \)
Given: \( 2x - 1 + 3x + 4 = 5x + 3 \)
Mistake: \( 2 \times (-1) \) should be \( -2 \), not \( -1 \); and \( 3 \times 4 \) should be \( 12 \), not \( 4 \).
Correct: \( 2(x - 1) + 3(x + 4) = 2x - 2 + 3x + 12 = 5x + 10 \)

Expression 3: \( y + 2(y + 2) \)
Given: \( (y + 2)^2 = y^2 + 4y + 4 \)
Mistake: \( y + 2(y + 2) \) is not the same as \( (y + 2)^2 \) - one is addition plus multiplication, the other is squaring.
Correct: \( y + 2(y + 2) = y + 2y + 4 = 3y + 4 \)

Expression 4: \( (5m + 6n)^2 \)
Given: \( 25m^2 + 36n^2 \)
Mistake: The middle term \( 2 \times 5m \times 6n \) is missing.
Correct: \( (5m + 6n)^2 = 25m^2 + 60mn + 36n^2 \)

Expression 5: \( (-q + 2)^2 \)
Given: \( q^2 - 4q + 4 \)
This is correct. No mistake.

Expression 6: \( 3a (2b \times 3c) \)
Given: \( 6ab \times 9ac = 54a^2bc \)
Mistake: The student multiplied in two separate stages when they should have simplified inside the bracket first and changed the expression incorrectly.
Correct way: \( 2b \times 3c = 6bc \); then \( 3a \times 6bc = 18abc \)
Correct final answer: \( 18abc \)

Expression 7: \( \frac{1}{2}(10s - 6) + 3 \)
Given: \( 5s - 3 + 3 = 5s \)
This is correct. No mistake.

Expression 8: \( 5w^2 + 6w \)
Given: \( 11w^2 \)
Mistake: \( w^2 \) and \( w \) are not like terms, so they cannot be added together.
Correct: Expression already in simplest form: \( 5w^2 + 6w \)

Expression 9: \( 2a^3 + 3a^3 + 6a^2b + 6ab^2 \)
Given: \( 5a^3 + 12a^2b^2 \)
Mistake: The student correctly added \( 2a^3 + 3a^3 = 5a^3 \), but then incorrectly added \( 6a^2b \) and \( 6ab^2 \) - these are not like terms (the powers of \( a \) and \( b \) are different), so they cannot be combined into \( 12a^2b^2 \).
Correct: \( 2a^3 + 3a^3 = 5a^3 \), so the final answer is: \( 5a^3 + 6a^2b + 6ab^2 \)

Expression 10: \( (x + 2)(x + 5) \)
Given: \( (x + 2)x + (x + 2)5 = x^2 + 2x + 5x + 10 = x^2 + 7x + 10 \)
This is correct. No mistake.

Expression 11: \( (a + 2)(b + 4) \)
Given: \( ab + 8 \)
Mistake: Only \( ab \) and \( 2 \times 4 \) are considered; the other products \( a \times 4 \) and \( 2 \times b \) are missing.
Correct: \( (a + 2)(b + 4) = ab + 4a + 2b + 8 \)

Expression 12: \( ab^2 + a^2b + a^2b^2 \)
Given: \( ab(a + b + ab) \)
Taking \( ab \) common:
\( ab^2 = ab \cdot b \)
\( a^2b = ab \cdot a \)
\( a^2b^2 = ab \cdot ab \)
So: \( ab^2 + a^2b + a^2b^2 = ab(b + a + ab) \)
And \( a + b = b + a \), so the factorisation is fine.
This is correct. No mistake.

In simple words: Always apply multiplication and distribution rules carefully - forgetting a term, confusing unlike terms, or mixing up operations like adding and multiplying are the most common mistakes.

Exam Tip: After expanding or simplifying, substitute a simple number (like x = 1, a = 2) into both your answer and the original expression to check if they are equal.

Question. Use this formula to find the number of circles in Step 15.
Answer: The number of circles in Step \( k \) is given by the formula: \( k^2 + 2k \)
Finding the number of circles in Step 15, substitute \( k = 15 \):
\( 15^2 + 2 \times 15 = 225 + 30 = 255 \)
Therefore, there are 255 circles in Step 15.
In simple words: Plug the step number into the formula like you would with any equation, and calculate the result.

Exam Tip: Always substitute the given value carefully and work through the arithmetic step-by-step to avoid calculation errors.

Question. Consider the pattern made of square tiles in the picture below.
Answer: The pattern is made from the square tiles: \( 3^2 - 1^2, 4^2 - 2^2, 5^2 - 3^2, \ldots \)
That is \( (k + 2)^2 - k^2 \)
So, the number of tiles in step \( k \): \( (k + 2)^2 - k^2 \)
\( = k^2 + 4k + 4 - k^2 \)
\( = 4k + 4 \)
Therefore:
In figure 1, \( k = 1 \): tiles = \( 4 \times 1 + 4 = 8 \)
In figure 2, \( k = 2 \): tiles = \( 4 \times 2 + 4 = 12 \)
In figure 3, \( k = 3 \): tiles = \( 4 \times 3 + 4 = 16 \)
In simple words: Recognize the difference-of-squares pattern in the figures, expand it using the identity, and you get a simple formula for any step number.

Exam Tip: Always try to spot algebraic patterns in geometric figures - this bridges visual and algebraic thinking.

Question. How many are there in Step 4 of the sequence? What about Step 10?
Answer: In step 4, \( k = 4 \):
\( \text{tiles} = 4 \times 4 + 4 = 20 \)
In step 10, \( k = 10 \):
\( \text{tiles} = 4 \times 10 + 4 = 44 \)
In simple words: Use the formula \( 4k + 4 \) by substituting the step number you want.

Exam Tip: Check your answer by counting or verifying with another method if possible - this confirms the formula is correct.

Question. Write an algebraic expression for the number of tiles in Step n. Share your methods with the class. Can you find more than one method to arrive at the answer?
Answer: The number of tiles in step \( n \): \( (n + 2)^2 - n^2 \)
\( = n^2 + 4n + 4 - n^2 \)
\( = 4n + 4 \)
In simple words: Replace the step number \( k \) with the letter \( n \) to get a general formula that works for any step.

Exam Tip: Creating a general formula from specific cases shows mastery of algebraic thinking - always try to do this in pattern questions.

Question. Find the area of the (interior) shaded region in the figure below. All four rectangles have the same dimensions.
Answer: Length of complete square region = \( n + m \)
So, the length of inner square region = \( n - m \)
Therefore, the area of inner square region = \( (n - m)^2 \).

By expanding both expressions, we check that \( (m + n)^2 - 4mn = (n - m)^2 \).

LHS = \( (m + n)^2 - 4mn \)
\( = (m^2 + 2mn + n^2) - 4mn \)
\( = m^2 + n^2 - 2mn \)

RHS = \( (n - m)^2 \)
\( = n^2 - 2nm + m^2 \)
\( = m^2 + n^2 - 2mn \)

So, LHS = RHS.
Therefore, \( (m + n)^2 - 4mn = (n - m)^2 \).
In simple words: The large square minus the four outer rectangles equals the inner square - by expanding both sides, you can see they are the same.

Exam Tip: When a visual geometry problem can be solved algebraically, always show both the geometric reasoning and the algebraic verification.

Question. Find out the area of the region with slanting lines in the figure. All three rectangles have the same dimensions (Fig. 1).
Answer: We have to find the area of the region with slanting lines.
All three rectangles in the figure have the same dimensions \( x \) by \( y \).
Area of ABCD = \( x^2 \).
Area of EFGH = \( xy \).
Required area = Area (ABCD) - Area (EFGH) = \( x^2 - xy \).
In simple words: The shaded region is the big square minus the inner rectangle.

Exam Tip: Identify the shapes involved (square, rectangle) and use their area formulas - breaking a complex region into simpler pieces is a powerful strategy.

Question. By expanding the expressions, verify that all three expressions are equivalent. If x = 8 and y = 3, find the area of the shaded region.
Answer: Expression 1: \( x^2 - xy \) (This is already simplified.)

Expression 2: \( x(x + 2y) - 3xy \)
\( = x^2 + 2xy - 3xy \)
\( = x^2 - xy \)
Matches Expression 1.

Expression 3: \( x(x - y) \)
\( = x^2 - xy \)
Matches Expression 1 and 2.

Therefore, all three expressions are equivalent and represent the same area.

Substituting \( x = 8 \) and \( y = 3 \) in area = \( x^2 - xy \):
Area = \( 8^2 - 8 \times 3 = 64 - 24 = 40 \)
In simple words: Different ways of writing the area formula all expand to the same result - this proves they are equivalent.

Exam Tip: Always expand each expression fully and simplify to show they match - this is more convincing than just looking at them.

Question. Write an expression for the area of the dashed region in the figure below. Use more than one method to arrive at the answer. Substitute p = 6, r = 3.5, and s = 9, and calculate the area.
Answer: Direct method:
Length of dashed region = \( (p - r) \)
Width of dashed region = \( (s - r) \)
So area = \( (p - r)(s - r) \)
\( = ps - pr - rs + r^2 \)

Rearranging the pieces:
Area of dashed region = area of complete rectangle - (area of two rectangles with length \( p \) and \( s \) having width \( r \)) + common area of two rectangles
\( = ps - (pr + sr) + r^2 \)
\( = ps - pr - sr + r^2 \)

Both methods give the same expression: \( ps - pr - sr + r^2 \)

Substituting \( p = 6, r = 3.5, s = 9 \):
Area = \( 6 \times 9 - 6 \times 3.5 - 3.5 \times 9 + 3.5^2 \)
\( = 54 - 21 - 31.5 + 12.25 \)
\( = 13.75 \)
In simple words: You can find the area by multiplying the dimensions directly, or by subtracting unwanted pieces from the whole - both give the same answer.

Exam Tip: Using multiple methods to solve the same problem builds confidence and provides a check on your answer.

Question 1. Compute these products using the suggested identity.
(i) 46² using Identity 1A for (a + b)²
(ii) 397 × 403 using Identity 1C for (a + b)(a - b)
(iii) 91² using Identity 1B for (a - b)²
(iv) 43 × 45 using Identity 1C for (a + b)(a - b)
Answer:
(i) Using Identity 1A: (a + b)² = a² + b² + 2ab
Start by writing 46 as 40 + 6.
Then (40 + 6)² = 40² + 6² + 2·40·6 = 1600 + 36 + 480 = 2116
So, 46² = 2116

(ii) Using Identity 1C: (a + b)(a - b) = a² - b²
Find the middle value: 400
Rewrite: 397 = 400 - 3 and 403 = 400 + 3
So, (400 - 3)(400 + 3) = 400² - 3² = 160000 - 9 = 159991
So, 397 × 403 = 159991

(iii) Using Identity 1B: (a - b)² = a² + b² - 2ab
Write 91 as 100 - 9.
Then (100 - 9)² = 100² + 9² - 2·100·9 = 10000 + 81 - 1800 = 8281
So, 91² = 8281

(iv) Using Identity 1C: (a + b)(a - b) = a² - b²
Find the middle value: 44
Rewrite: 43 = 44 - 1 and 45 = 44 + 1
Then (44 - 1)(44 + 1) = 44² - 1² = 1936 - 1 = 1935
So, 43 × 45 = 1935
In simple words: Break each number into a middle value plus or minus a small part. Use the right identity formula to make the calculation quick and easy.

Exam Tip: Always identify which identity applies before starting. The middle value method saves a lot of arithmetic and cuts down errors.

Question 2. Use either a suitable identity or the distributive property to find each of the following products.
(i) (p - 1)(p + 11)
(ii) (3a - 9b)(3a + 9b)
(iii) -(2y + 5)(3y + 4)
(iv) (6x + 5y)²
(v) (2x - 1/2)²
(vi) (7p) × (3r) × (p + 2)
Answer:
(i) Apply the distributive property: (p - 1)(p + 11) = p(p + 11) - 1(p + 11) = (p² + 11p) - (p + 11) = p² + 10p - 11

(ii) Notice this has the form (A - B)(A + B) = A² - B² where A = 3a and B = 9b.
(3a - 9b)(3a + 9b) = (3a)² - (9b)² = 9a² - 81b²

(iii) First expand (2y + 5)(3y + 4) by distributing: 2y·3y + 2y·4 + 5·3y + 5·4 = 6y² + 8y + 15y + 20 = 6y² + 23y + 20
Then apply the minus sign: -(6y² + 23y + 20) = -6y² - 23y - 20

(iv) Use (a + b)² = a² + b² + 2ab
(6x + 5y)² = (6x)² + (5y)² + 2·6x·5y = 36x² + 25y² + 60xy

(v) Use (a - b)² = a² + b² - 2ab
(2x - 1/2)² = (2x)² + (1/2)² - 2·2x·(1/2) = 4x² + 1/4 - 2x

(vi) Combine the constants first: 7p × 3r = 21pr
Then distribute: 21pr(p + 2) = 21pr·p + 21pr·2 = 21p²r + 42pr
In simple words: Pick the right method - either a quick identity pattern or spreading the terms across. Combine like parts at the end.

Exam Tip: Watch for patterns that match (a + b)², (a - b)², or difference of squares before using the long distributive method.

Question 3. For each statement identify the appropriate algebraic expression(s).
(i) Two more than a square number.
Answer: s² + 2
In simple words: Take any number s, square it, then add 2 to get the final result.

Exam Tip: Always translate words into symbols in the right order - "two more" means add 2 to whatever comes before it.

Question 3. (ii) The sum of the squares of two consecutive numbers.
Answer: m² + (m + 1)²
In simple words: Use one number as m. The next number is m + 1. Square both and add them together.

Exam Tip: For consecutive numbers, always name the first one m, so the second is automatically m + 1. This keeps the expression clean.

Question 4. Consider any 2 by 2 square of numbers in a calendar, as shown in the figure. Find products of numbers lying along each diagonal - 4 × 12 = 48, 5 × 11 = 55. Do this for the other 2 by 2 squares. What do you observe about the diagonal products? Explain why this happens.
Answer: Take a 2 by 2 square from any calendar. Compute the two diagonal products.

Example: From the given square, 7 × 15 = 105 and 8 × 14 = 112. The difference is 112 - 105 = 7.

Testing other 2×2 squares gives the same outcome: one diagonal product is always 7 more than the other.

To see why, label any 2×2 block like this:
top-left: a
top-right: a + 1
bottom-left: a + 7
bottom-right: a + 8

Diagonal 1: a(a + 8) = a² + 8a
Diagonal 2: (a + 1)(a + 7) = a² + 7a + a + 7 = a² + 8a + 7

The difference is (a² + 8a + 7) - (a² + 8a) = 7

This proves that for every 2×2 calendar square, one diagonal product is always 7 more than the other.
In simple words: Pick any four dates in a 2-by-2 box from a calendar. Multiply the two corners going one way, then the other way. The answers are always 7 apart.

Exam Tip: Use variables to prove patterns that work for any case, not just one example. The proof shows it works forever.

Question 5. Verify which of the following statements are true.
(i) (k + 1)(k + 2) - (k + 3) is always 2.
(ii) (2q + 1)(2q - 3) is a multiple of 4.
(iii) Squares of even numbers are multiples of 4, and squares of odd numbers are 1 more than multiples of 8.
(iv) (6n + 2)² - (4n + 3)² is 5 less than a square number.
Answer:
(i) Expand (k + 1)(k + 2) = k² + 3k + 2
Subtract (k + 3): (k² + 3k + 2) - (k + 3) = k² + 2k - 1
Test with values: If k = 1, get 1 + 2 - 1 = 2 ✓. If k = 2, get 4 + 4 - 1 = 7 ✗
This is not always 2, so the statement is FALSE.

(ii) Expand (2q + 1)(2q - 3) = 4q² - 6q + 2q - 3 = 4q² - 4q - 3
Rewrite as 4(q² - q) - 3
Since we have a -3 at the end, this is 3 less than a multiple of 4, not a multiple of 4 itself. The statement is FALSE.

(iii) For an even number 2m: (2m)² = 4m², which is a multiple of 4 ✓
For an odd number 2m + 1: (2m + 1)² = 4m² + 4m + 1 = 4m(m + 1) + 1
Since m(m + 1) is always even (product of consecutive integers), write it as 2k. Then 4·2k + 1 = 8k + 1, which is 1 more than a multiple of 8 ✓
The statement is TRUE.

(iv) Using the difference of squares: (6n + 2)² - (4n + 3)² = [(6n + 2) + (4n + 3)][(6n + 2) - (4n + 3)]
= (10n + 5)(2n - 1)
= 5(2n + 1)(2n - 1)
= 5(4n² - 1)
= 20n² - 5
This can be written as (10n)² - 5, which is 5 less than the square (10n)². The statement is TRUE.
In simple words: Test each claim by expanding, simplifying, or plugging in numbers. Some are always true, others fail with just one example.

Exam Tip: A single counterexample kills a "always true" claim. Algebraic proof shows something works forever.

Question 6. A number leaves a remainder of 3 when divided by 7, and another number leaves a remainder of 5 when divided by 7. What is the remainder when their sum, difference, and product are divided by 7?
Answer: Let the first number be 7a + 3 (remainder 3 when divided by 7)
Let the second number be 7b + 5 (remainder 5 when divided by 7)

For the sum:
(7a + 3) + (7b + 5) = 7a + 7b + 8 = 7(a + b) + 8
Divide 8 by 7: 8 = 7·1 + 1, so remainder is 1
Remainder of the sum = 1

For the difference:
(7a + 3) - (7b + 5) = 7a - 7b - 2 = 7(a - b) - 2
For -2 divided by 7: -2 + 7 = 5, so remainder is 5
Remainder of the difference = 5

For the product:
(7a + 3)(7b + 5) = 49ab + 35a + 21b + 15 = 7(7ab + 5a + 3b) + 15
Divide 15 by 7: 15 = 7·2 + 1, so remainder is 1
Remainder of the product = 1
In simple words: When dividing by 7, ignore the multiple of 7 in each number and work only with the remainders: 3 and 5. Then find remainders of the sum, difference, and product.

Exam Tip: For negative remainders like -2, add 7 to get the positive equivalent (5 in this case). Always give remainders as positive numbers from 0 to 6.

Question 7. Choose three consecutive numbers, square the middle one, and subtract the product of the other two. Repeat the same with other sets of numbers. What pattern do you notice? How do we write this as an algebraic equation? Expand both sides of the equation to check that it is a true identity.
Answer: Test with examples:
Numbers 2, 3, 4: 3² - (2 × 4) = 9 - 8 = 1
Numbers 5, 6, 7: 6² - (5 × 7) = 36 - 35 = 1
Numbers 10, 11, 12: 11² - (10 × 12) = 121 - 120 = 1

The pattern: For any three consecutive numbers, the square of the middle number minus the product of the other two always equals 1.

Algebraic form: Let the three consecutive numbers be n - 1, n, and n + 1
Then: n² - (n - 1)(n + 1) = 1

Verification by expanding:
Expand (n - 1)(n + 1) = n² - 1
Substitute: n² - (n² - 1) = n² - n² + 1 = 1 ✓

The left side simplifies to 1, which equals the right side. Therefore, this is a true identity for any three consecutive numbers.
In simple words: Pick any three numbers in a row. Square the middle one. Multiply the other two. The first answer is always 1 more than the second.

Exam Tip: Expanding and verifying both sides proves an identity works for all numbers, not just the ones you tested.

Question 8. What is the algebraic expression describing the following steps - add any two numbers. Multiply this by half of the sum of the two numbers? Prove that this result will be half of the square of the sum of the two numbers.
Answer: Let the two numbers be a and b.
Sum = a + b
Half of the sum = 1/2(a + b)
Multiply the sum by half the sum: (a + b) × 1/2(a + b)

The algebraic expression is: (a + b) × 1/2(a + b)

To find half of the square of the sum:
Square of the sum = (a + b)²
Half of that = 1/2(a + b)²

Now expand the expression (a + b) × 1/2(a + b):
= 1/2(a + b)(a + b)
= 1/2(a + b)²

This is exactly the same as half of the square of the sum. Therefore, the result is always half of the square of the sum of the two numbers.
In simple words: Add two numbers. Take half of that sum and multiply it by the whole sum. You get exactly the same as taking half of what you'd get if you squared the sum.

Exam Tip: Algebra helps prove relationships that seem mysterious with just numbers. The expansion shows why the two approaches give identical results.

Question 9. Which is larger? Find out without fully computing the product.
(i) 14 × 26 or 16 × 24
(ii) 25 × 75 or 26 × 74
Answer:
(i) Check the sums:
14 + 26 = 40
16 + 24 = 40

Both pairs have the same sum. When the sum is fixed, the product gets bigger as the two numbers get closer to each other.
14 and 26 are far apart (difference = 12)
16 and 24 are closer (difference = 8)
Therefore, 16 × 24 is larger.

(ii) Check the sums:
25 + 75 = 100
26 + 74 = 100

Both sums are the same. Compare closeness:
25 and 75 have a difference of 50
26 and 74 have a difference of 48
Since 26 and 74 are closer, 26 × 74 is the larger product.
In simple words: If two pairs of numbers add up to the same total, the pair that is closer together will have a bigger product.

Exam Tip: Always check if sums are equal before comparing products - this rule only works when the sum is the same for both pairs.

Question 10. A tiny park is coming up in Dhauli. The plan is shown in the figure. The two square plots, each of area g² sq. ft., will have a green cover. All the remaining area is a walking path w ft. wide that needs to be tiled. Write an expression for the area that needs to be tiled.
Answer: Each green square has area g² sq. ft., so each side is g ft.

The walking path is w ft. wide on the left, right, top, and bottom. Between the two squares, the path is 2w ft. wide.

For the outer rectangle dimensions:
Total width = g (height of square) + w (top path) + w (bottom path) = g + 2w
Total length = w (left path) + g (first square) + 2w (path between squares) + g (second square) + w (right path) = 2g + 4w

Total area of the park = (g + 2w)(2g + 4w)
Green area = 2g²

Area to be tiled (walking path) = (g + 2w)(2g + 4w) - 2g²
In simple words: Find the whole park size by length times width. Subtract the two green square areas. What is left is the path that needs tiles.

Exam Tip: Draw and label all dimensions carefully. The gap between the two squares is 2w, not w - count it correctly.

Question 11. For each pattern shown below:
(i) Draw the next figure in the sequence.
Answer: For the first pattern (yellow squares in an L-shape): The sequence grows by adding one more row to the vertical arm and one more column to the horizontal arm. Step 1 has 5 squares, Step 2 has 9 squares, Step 3 has 13 squares. Step 4 continues this growth with squares arranged in an expanded L-shape.

For the second pattern (blue squares in a grid): The sequence builds rectangular grids. Step 1 is 2 × 2 plus 1 extra = 5 squares. Step 2 is 3 × 3 = 9 squares. Step 3 is 4 × 4 = 16 squares. Step 4 is 5 × 5 = 25 squares.
In simple words: In the first pattern, you add more steps to both arms of the L. In the second pattern, each step is a bigger square grid.

Exam Tip: Count carefully how many squares are added each time. Look for whether the growth is by rows, columns, or both together.

Question 11. (ii) How many basic units are there in Step 10?
Answer: For the first pattern (L-shaped), the rule is (k + 2)² where k is the step number.
Step 1: k = 1, so (1 + 2)² = 9 squares
Step 2: k = 2, so (2 + 2)² = 16 squares
Step 3: k = 3, so (3 + 2)² = 25 squares
Step 4: k = 4, so (4 + 2)² = 36 squares
Step 10: k = 10, so (10 + 2)² = 144 squares

For the second pattern (grid-based), the rule is (k + 1)² + k where k is the step number.
Step 1: k = 1, so (1 + 1)² + 1 = 4 + 1 = 5 squares
Step 2: k = 2, so (2 + 1)² + 2 = 9 + 2 = 11 squares
Step 3: k = 3, so (3 + 1)² + 3 = 16 + 3 = 19 squares
Step 4: k = 4, so (4 + 1)² + 4 = 25 + 4 = 29 squares
Step 10: k = 10, so (10 + 1)² + 10 = 121 + 10 = 131 squares
In simple words: Find the pattern rule. Plug in 10 for the step number. Calculate to get the answer.

Exam Tip: Look for what changes by the same amount each step - this helps you find the algebraic rule. Then use that rule for any step.

Question 11. (iii) Write an expression to describe the number of basic units in Step y.
Answer: For the first pattern (L-shaped figures): Step y = (y + 2)²

For the second pattern (grid patterns): Step y = (y + 1)² + y
In simple words: Replace the step number k with y in the formula you found. That formula now works for any step.

Exam Tip: Always verify your formula by testing it with a known step (like Step 1 or Step 2) before declaring it correct.