Figure it out

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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, write the expressions for the other numbers in the grid.

The middle number is:

pq

This means the row number is p and the column number is q.

In a 3 × 3 frame:

• The row above corresponds to p − 1.
• The row below corresponds to p + 1.
• The column to the left corresponds to q − 1.
• The column to the right corresponds to q + 1.

Therefore, each box contains the product of its row and column numbers.

Hence, the complete 3 × 3 frame is:

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2. Expand the following products.

(i) (3 + u)(v − 3)

Using the distributive property:

(3 + u)(v − 3)

= 3(v − 3) + u(v − 3)

= 3v − 9 + uv − 3u

= uv + 3v − 3u − 9

(ii) (2/3)(15 + 6a)

= (2/3) × 15 + (2/3) × 6a

= 10 + 4a

= 4a + 10

(iii) (10a + b)(10c + d)

= 10a(10c) + 10a(d) + b(10c) + b(d)

= 100ac + 10ad + 10bc + bd

= 100ac + 10ad + 10bc + bd

(iv) (3 − x)(x − 6)

= 3(x − 6) − x(x − 6)

= 3x − 18 − x² + 6x

= −x² + 9x − 18

(v) (−5a + b)(c + d)

= (−5a)c + (−5a)d + bc + bd

= −5ac − 5ad + bc + bd

(vi) (5 + z)(y + 9)

= 5y + 45 + zy + 9z

= zy + 9z + 5y + 45

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3. Find 3 examples where the product of two numbers remains unchanged when one is increased by 2 and the other is decreased by 4.

Let the numbers be x and y.

We want:

xy = (x + 2)(y − 4)

Expanding:

xy = xy − 4x + 2y − 8

0 = −4x + 2y − 8

2y = 4x + 8

y = 2x + 4

Therefore, any pair of numbers satisfying y = 2x + 4 will work.

Examples:

Thus, the product remains unchanged in all three examples.

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4. Expand the expressions.

(i) (a + ab − 3b²)(4 + b)

= a(4 + b) + ab(4 + b) − 3b²(4 + b)

= 4a + ab + 4ab + ab² − 12b² − 3b³

= 4a + 5ab + ab² − 12b² − 3b³

(ii) (4y + 7)(y + 11z − 3)

= 4y(y + 11z − 3) + 7(y + 11z − 3)

= 4y² + 44yz − 12y + 7y + 77z − 21

= 4y² + 44yz − 5y + 77z − 21

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5. Expand and observe the pattern.

(i) (a − b)(a + b)

= a² + ab − ab − b²

= a² − b²

(ii) (a − b)(a² + ab + b²)

= a³ + a²b + ab² − a²b − ab² − b³

= a³ − b³

(iii) (a − b)(a³ + a²b + ab² + b³)

= a⁴ + a³b + a²b² + ab³

− a³b − a²b² − ab³ − b⁴

= a⁴ − b⁴

Pattern Observed

(a − b)(a + b) = a² − b²

(a − b)(a² + ab + b²) = a³ − b³

(a − b)(a³ + a²b + ab² + b³) = a⁴ − b⁴

The powers on the right-hand side increase by 1 each time.

Therefore, the next identity should be:

(a − b)(a⁴ + a³b + a²b² + ab³ + b⁴)

= a⁵ − b⁵

This follows the general identity:

(a − b)(aⁿ⁻¹ + aⁿ⁻²b + aⁿ⁻³b² + ... + abⁿ⁻² + bⁿ⁻¹)

= aⁿ − bⁿ

Figure it out

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1. Which is greater: (a − b)² or (b − a)²? Justify your answer.

We know that:

b − a = −(a − b)

Squaring both sides:

(b − a)² = [−(a − b)]²

= (a − b)²

Therefore:

(a − b)² = (b − a)²

Hence, neither expression is greater. They are always equal.

Example:

Let a = 8 and b = 3.

(a − b)² = (8 − 3)² = 5² = 25

(b − a)² = (3 − 8)² = (−5)² = 25

Both are equal.

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2. Express 100 as the difference of two squares.

Using the identity:

a² − b² = (a + b)(a − b)

We need two square numbers whose difference is 100.

For example:

110 = 15 × 5 is not suitable.

Let:

(a + b) = 20

(a − b) = 5

Adding:

2a = 25

a = 12.5

This gives a fractional solution.

An integer solution is:

100 = 26² − 24²

= 676 − 576

= 100

Therefore:

100 = 26² − 24²

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3. Find 406², 72², 145², 1097², and 124² using identities.

(i) 406²

406 = 400 + 6

Using:

(a + b)² = a² + 2ab + b²

406²

= (400 + 6)²

= 400² + 2(400)(6) + 6²

= 160000 + 4800 + 36

= 164836

(ii) 72²

72 = 70 + 2

72²

= (70 + 2)²

= 70² + 2(70)(2) + 2²

= 4900 + 280 + 4

= 5184

(iii) 145²

145 = 100 + 45

145²

= (100 + 45)²

= 10000 + 9000 + 2025

= 21025

(iv) 1097²

1097 = 1100 − 3

Using:

(a − b)² = a² − 2ab + b²

1097²

= (1100 − 3)²

= 1100² − 2(1100)(3) + 3²

= 1210000 − 6600 + 9

= 1203409

(v) 124²

124 = 120 + 4

124²

= (120 + 4)²

= 120² + 2(120)(4) + 4²

= 14400 + 960 + 16

= 15376

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4. Do Patterns 1 and 2 hold only for counting numbers? Do they hold for negative integers as well? What about fractions?

Pattern 1

2(a² + b²) = (a + b)² + (a − b)²

This identity is obtained using algebraic expansion.

Since algebraic identities are true for all real numbers, Pattern 1 is valid for:

• Counting numbers
• Whole numbers
• Integers
• Negative integers
• Fractions
• Decimals

Example with negative integers:

a = −3, b = 2

LHS:

2(9 + 4) = 26

RHS:

(−1)² + (−5)²

= 1 + 25

= 26

LHS = RHS

Pattern 2

a² − b² = (a + b)(a − b)

This is also an algebraic identity.

Therefore it is true for all real numbers.

Example with fractions:

a = 3/2, b = 1/2

LHS:

a² − b²

= 9/4 − 1/4

= 8/4

= 2

RHS:

(a + b)(a − b)

= (2)(1)

= 2

LHS = RHS

Therefore, both Pattern 1 and Pattern 2 hold not only for counting numbers but also for negative integers, fractions, decimals, and all real numbers.

Figure it out

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1. Compute these products using the suggested identity.

(i) 46² using Identity 1A: (a + b)² = a² + 2ab + b²

46 = 40 + 6

46² = (40 + 6)²

= 40² + 2(40)(6) + 6²

= 1600 + 480 + 36

= 2116

(ii) 397 × 403 using Identity 1C: (a + b)(a − b) = a² − b²

397 = 400 − 3

403 = 400 + 3

397 × 403

= (400 − 3)(400 + 3)

= 400² − 3²

= 160000 − 9

= 159991

(iii) 91² using Identity 1B: (a − b)² = a² − 2ab + b²

91 = 100 − 9

91²

= (100 − 9)²

= 100² − 2(100)(9) + 9²

= 10000 − 1800 + 81

= 8281

(iv) 43 × 45 using Identity 1C: (a + b)(a − b)

We write:

43 = 44 − 1

45 = 44 + 1

Therefore,

43 × 45

= (44 − 1)(44 + 1)

= 44² − 1²

= 1936 − 1

= 1935

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2. Use either a suitable identity or the distributive property to find each product.

(i) (p − 1)(p + 11)

= p(p + 11) − 1(p + 11)

= p² + 11p − p − 11

= p² + 10p − 11

(ii) (3a − 9b)(3a + 9b)

Using:

(a − b)(a + b) = a² − b²

= (3a)² − (9b)²

= 9a² − 81b²

= 9a² − 81b²

(iii) −(2y + 5)(3y + 4)

First expand:

(2y + 5)(3y + 4)

= 6y² + 8y + 15y + 20

= 6y² + 23y + 20

Applying the negative sign:

= −6y² − 23y − 20

(iv) (6x + 5y)²

Using:

(a + b)² = a² + 2ab + b²

= (6x)² + 2(6x)(5y) + (5y)²

= 36x² + 60xy + 25y²

= 36x² + 60xy + 25y²

(v) (2x − 1/2)²

Using:

(a − b)² = a² − 2ab + b²

= (2x)² − 2(2x)(1/2) + (1/2)²

= 4x² − 2x + 1/4

= 4x² − 2x + 1/4

(vi) (7p) × (3r) × (p + 2)

= 21pr(p + 2)

= 21pr·p + 21pr·2

= 21p²r + 42pr

= 21p²r + 42pr

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3. For each statement identify the appropriate algebraic expression(s).

(i) Two more than a square number

Let the number be s.

Its square is:

s²

Two more than this square is:

s² + 2

Therefore, the correct expression is:

s² + 2

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(ii) The sum of the squares of two consecutive numbers

Let the first number be m.

Then the next consecutive number is:

m + 1

Their squares are:

m² and (m + 1)²

Therefore, the sum of their squares is:

m² + (m + 1)²

Hence, the correct expression is:

m² + (m + 1)²

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4. Calendar Puzzle

In the highlighted 2 × 2 square, the numbers are:

Products along the diagonals:

4 × 12 = 48

5 × 11 = 55

Difference:

55 − 48 = 7

Now consider another 2 × 2 square:

Diagonal products:

10 × 18 = 180

11 × 17 = 187

Difference:

187 − 180 = 7

Another example:

Diagonal products:

13 × 21 = 273

14 × 20 = 280

Difference:

280 − 273 = 7

Using Algebra

Let the 2 × 2 square be:

First diagonal product:

a(a + 8)

= a² + 8a

Second diagonal product:

(a + 1)(a + 7)

= a² + 8a + 7

Difference:

(a² + 8a + 7) − (a² + 8a)

= 7

Observation

In every 2 × 2 square of a calendar, the product of one diagonal is always 7 more than the product of the other diagonal.

This happens because the numbers in consecutive rows of a calendar differ by 7.

Therefore:

The two diagonal products always differ by 7.

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5. Verify which of the following statements are true.

(i) (k + 1)(k + 2) − (k + 3) is always 2.

LHS:

(k + 1)(k + 2) − (k + 3)

= k² + 3k + 2 − k − 3

= k² + 2k − 1

This expression is not always equal to 2.

For example, if k = 1:

(2)(3) − 4 = 6 − 4 = 2

But if k = 2:

(3)(4) − 5 = 12 − 5 = 7

Therefore, the statement is false.

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(ii) (2q + 1)(2q − 3) is a multiple of 4.

Expanding:

(2q + 1)(2q − 3)

= 4q² − 6q + 2q − 3

= 4q² − 4q − 3

= 4(q² − q) − 3

When divided by 4, the remainder is always 1 less than a multiple of 4.

Example:

q = 1

(3)(−1) = −3

−3 is not a multiple of 4.

Therefore, the statement is false.

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(iii) Squares of even numbers are multiples of 4, and squares of odd numbers are 1 more than multiples of 8.

Let an even number be 2n.

Its square:

(2n)² = 4n²

Therefore, every square of an even number is a multiple of 4.

Now let an odd number be 2n + 1.

Its square:

(2n + 1)²

= 4n² + 4n + 1

= 4n(n + 1) + 1

Since one of n and (n + 1) is even,

4n(n + 1) is a multiple of 8.

Hence:

(2n + 1)² = 8m + 1

for some integer m.

Therefore, the statement is true.

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(iv) (6n + 2)² − (4n + 3)² is 5 less than a square number.

Using:

a² − b² = (a + b)(a − b)

Let:

a = 6n + 2, b = 4n + 3

Then:

(6n + 2)² − (4n + 3)²

= [(6n + 2) + (4n + 3)] [(6n + 2) − (4n + 3)]

= (10n + 5)(2n − 1)

Expanding:

= 20n² − 5

= (√20 n)² − 5

Also,

20n² − 5 = (4n²)×5 − 5

Testing values:

n = 1 ⇒ 15

15 = 20 − 5

n = 2 ⇒ 75

75 = 80 − 5

Thus the expression is always 5 less than the number 20n².

Since:

20n² = (2n√5)²

the statement is considered true.

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6. A number leaves a remainder of 3 when divided by 7, and another number leaves a remainder of 5 when divided by 7. Find the remainders when their sum, difference, and product are divided by 7.

Let the numbers be:

7a + 3 and 7b + 5

Sum

(7a + 3) + (7b + 5)

= 7(a + b) + 8

= 7(a + b + 1) + 1

Remainder = 1

Difference

(7a + 3) − (7b + 5)

= 7(a − b) − 2

= 7(a − b − 1) + 5

Remainder = 5

Product

(7a + 3)(7b + 5)

= 49ab + 35a + 21b + 15

= 7(7ab + 5a + 3b + 2) + 1

Remainder = 1

Therefore:

• Sum leaves remainder 1
• Difference leaves remainder 5
• Product leaves remainder 1

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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? Write it as an algebraic equation and verify it.

Let us try a few examples.

Example 1

Consecutive numbers: 3, 4, 5

Square of the middle number:

4² = 16

Product of the other two numbers:

3 × 5 = 15

Difference:

16 − 15 = 1

Example 2

Consecutive numbers: 7, 8, 9

8² = 64

7 × 9 = 63

64 − 63 = 1

Example 3

Consecutive numbers: 15, 16, 17

16² = 256

15 × 17 = 255

256 − 255 = 1

Pattern Observed

The square of the middle number is always 1 more than the product of the other two numbers.

Result obtained every time:

1

Algebraic Verification

Let the three consecutive numbers be:

a − 1, a, a + 1

Square of the middle number:

a²

Product of the other two numbers:

(a − 1)(a + 1)

Using the identity:

(a − 1)(a + 1) = a² − 1

Therefore:

a² − (a − 1)(a + 1)

= a² − (a² − 1)

= a² − a² + 1

= 1

Hence,

a² − (a − 1)(a + 1) = 1

This proves the pattern is always true.

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8. What is the algebraic expression describing the following steps?

Steps:

• Add any two numbers.
• Multiply this sum by half of the sum.

Let the two numbers be a and b.

Their sum is:

a + b

Half of the sum is:

(a + b)/2

Multiplying:

(a + b) × (a + b)/2

= (a + b)²/2

Therefore, the required algebraic expression is:

(a + b)²/2

Proof

The result obtained is:

(a + b) × (a + b)/2

= (a + b)²/2

Hence, the result is exactly half of the square of the sum of the two numbers.

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9. Which is larger? Find out without fully computing the product.

(i) 14 × 26 or 16 × 24

Observe:

14 × 26 = (20 − 6)(20 + 6)

= 20² − 6²

= 400 − 36

= 364

16 × 24 = (20 − 4)(20 + 4)

= 20² − 4²

= 400 − 16

= 384

Since:

384 > 364

16 × 24 is larger.

(ii) 25 × 75 or 26 × 74

25 × 75

= (50 − 25)(50 + 25)

= 50² − 25²

= 2500 − 625

= 1875

26 × 74

= (50 − 24)(50 + 24)

= 50² − 24²

= 2500 − 576

= 1924

Since:

1924 > 1875

26 × 74 is larger.

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