We Distribute, Yet Things Multiply

6.1 Some Properties of Multiplication

Distributivity: a(b+c)=ab+ac. For any a,b,m,n, (a+m)(b+n)=ab+an+bm+mn. The change from ab is an+bm+mn. If a increases by 1 and b decreases by 1, the change is b−a−1. For 23×27, increasing both by one adds 23+27+1=51. A rectangle model partitions the total area into four regions ab,an,bm,mn.

Multiplying expressions and like terms

Expand every term of one bracket against every term of the other, preserving signs. Example: (2x−3)(x+4)=2x²+8x−3x−12=2x²+5x−12. Like terms have identical variable parts and exponents; 3x² and 5x are unlike.

6.2 Special Cases of Distributivity

The identities (a+b)²=a²+2ab+b², (a−b)²=a²−2ab+b² and (a+b)(a−b)=a²−b² follow from multiplication. Example: 65²=(60+5)²=3600+600+25=4225. The middle term 2ab is essential. (a−b)² is not a²−b².

Fast multiplication with 11, 101, 1001 and 99

Use n×11=10n+n, n×101=100n+n, n×1001=1000n+n, n×99=100n−n. Examples: 94×11=1034; 495×11=5445; 89×101=8989; 9734×99=963666. Handle carries carefully.

6.3 Mind the Mistake, Mend the Mistake

To audit an expansion, distribute to every term, multiply signs, and combine only like terms. Examples: −3p(−5p+2q)=15p²−6pq; 2(x−1)+3(x+4)=5x+10; (5m+6n)²=25m²+60mn+36n²; (−q+2)²=q²−4q+4; (a+2)(b+4)=ab+4a+2b+8. Check by substituting simple values.

6.4 This Way or That Way, All Ways Lead to the Bay

One dot pattern can be described by (k+1)²−1, k²+2k, k(k+1)+k or k(k+2). Expanding proves they are equivalent. At Step 10, 10²+20=120 dots; at Step 15, 225+30=255. A square-tile border can be counted as (n+2)²−n²=4n+4. Area models help prove algebraic identities geometrically.

Comparison of identities

Original exam practice, MCQs and competency

MCQ: Expand (x+5)²: answer x²+10x+25. HOTS: Find 1002² mentally: (1000+2)²=1004004. Case study: A garden 12 m by 8 m is extended by 2 m on each dimension. New area=14×10=140 m², old area=96 m², increase=44 m². Competency: A shop sells 99 items at ₹47 each; use (100−1)×47=4653. These are original questions, not textbook exercises.

Flashcards, quick revision and common mistakes

Flashcard: distributive property? Multiply every term inside brackets. Flashcard: square of sum? a²+2ab+b². Flashcard: difference of squares? (a+b)(a−b). Flashcard: why combine like terms? Only matching variable factors represent the same quantity. Common mistakes: missing a cross-term, incorrect minus signs, confusing a²+b² with (a+b)², and treating a diagram as a proof without explaining its area partition.

NCERT Figure it Out — textbook pages 142–143

Question statements are paraphrased from the official NCERT 2026–27 reprint. Every numbered question has a separate explanation and final answer, including labelled subparts.

1. Question 1. Complete the 3×3 multiplication grid when its centre is pq.

   Step-by-step solution: Use row factors p−1,p,p+1 and column factors q−1,q,q+1. Multiply each row and column label.

   Final answer:

   (p−1)(q−1)	(p−1)q	(p−1)(q+1)
   p(q−1)	pq	p(q+1)
   (p+1)(q−1)	(p+1)q	(p+1)(q+1)

2. Question 2. Expand the six given products using distributivity.

   Step-by-step solution: (i) (3+u)(v−3)=3v−9+uv−3u. (ii) (2/3)(15+6a)=10+4a. (iii) (10a+b)(10c+d)=100ac+10ad+10bc+bd. (iv) (3−x)(x−6)=3x−18−x²+6x. (v) (−5a+b)(c+d)=−5ac−5ad+bc+bd. (vi) (5+z)(y+9)=5y+45+zy+9z.

   Final answer: (i) uv+3v−3u−9; (ii) 10+4a; (iii) 100ac+10ad+10bc+bd; (iv) −x²+9x−18; (v) −5ac−5ad+bc+bd; (vi) zy+5y+9z+45

3. Question 3. Find three pairs whose product remains unchanged after adding 2 to the first and subtracting 4 from the second.

   Step-by-step solution: Set (a+2)(b−4)=ab. Expand: ab−4a+2b−8=ab; hence b=2a+4. Choose a=1,2,3.

   Final answer: (1,6), (2,8), (3,10)

4. Question 4. Expand the two multivariable expressions.

   Step-by-step solution: (i) (a+ab−3b²)(4+b)=4a+4ab−12b²+ab+ab²−3b³. Combine 4ab+ab. (ii) (4y+7)(y+11z−3)=4y²+44yz−12y+7y+77z−21.

   Final answer: (i) 4a+5ab+ab²−12b²−3b³; (ii) 4y²+44yz−5y+77z−21

5. Question 5. Expand three difference products and identify the next identity.

   Step-by-step solution: (i) (a−b)(a+b)=a²−b². (ii) (a−b)(a²+ab+b²)=a³−b³. (iii) (a−b)(a³+a²b+ab²+b³)=a⁴−b⁴. The intermediate terms cancel pairwise.

   Final answer: Next: (a−b)(a⁴+a³b+a²b²+ab³+b⁴)=a⁵−b⁵

NCERT Figure it Out — textbook page 149

Question statements are paraphrased from the official NCERT 2026–27 reprint. Every numbered question has a separate explanation and final answer, including labelled subparts.

1. Question 1. Which is larger: (a−b)² or (b−a)²?

   Step-by-step solution: Since b−a=−(a−b), squaring removes the sign: [−(a−b)]²=(a−b)².

   Final answer: They are equal for all real a,b.

2. Question 2. Express 100 as a difference of two squares.

   Step-by-step solution: Use x²−y²=(x−y)(x+y). Choose x−y=2 and x+y=50, giving x=26,y=24.

   Final answer: 100=26²−24²=676−576

3. Question 3. Calculate 406², 72², 145², 1097² and 124² using identities.

   Step-by-step solution: 406²=(400+6)²=160000+4800+36=164836; 72²=(70+2)²=4900+280+4=5184; 145²=(100+45)²=10000+9000+2025=21025; 1097²=(1100−3)²=1210000−6600+9=1203409; 124²=(120+4)²=14400+960+16=15376.

   Final answer: 164836; 5184; 21025; 1203409; 15376

4. Question 4. Do the two algebraic patterns work for negative numbers and fractions?

   Step-by-step solution: Both patterns are polynomial identities derived by distributivity: (a+b)²+(a−b)²=2(a²+b²), and a²−b²=(a+b)(a−b). Polynomial identities hold for all real a,b, including negative numbers and fractions.

   Final answer: Yes, both identities hold for negative numbers and fractions.

NCERT Figure it Out — textbook pages 154–156

Question statements are paraphrased from the official NCERT 2026–27 reprint. Every numbered question has a separate explanation and final answer, including labelled subparts.

1. Question 1. Use the suggested identities to calculate four products.

   Step-by-step solution: (i) 46²=(40+6)²=1600+480+36. (ii) 397×403=(400−3)(400+3)=160000−9. (iii) 91²=(100−9)²=10000−1800+81. (iv) 43×45=(44−1)(44+1)=1936−1.

   Final answer: (i) 2116; (ii) 159991; (iii) 8281; (iv) 1935

2. Question 2. Expand six expressions using distributivity or an identity.

   Step-by-step solution: (i) (p−1)(p+11)=p²+11p−p−11. (ii) (3a−9b)(3a+9b)=9a²−81b². (iii) −(2y+5)(3y+4)=−(6y²+23y+20). (iv) (6x+5y)²=36x²+60xy+25y². (v) (2x−1/2)²=4x²−2x+1/4. (vi) (7p)(3r)(p+2)=21pr(p+2).

   Final answer: (i) p²+10p−11; (ii) 9a²−81b²; (iii) −6y²−23y−20; (iv) 36x²+60xy+25y²; (v) 4x²−2x+1/4; (vi) 21p²r+42pr

3. Question 3. Identify expressions for two more than a square and the sum of squares of consecutive numbers.

   Step-by-step solution: (i) A square number is s², so two more is s²+2. (ii) Consecutive integers are m and m+1, so their square sum is m²+(m+1)²; m²+(m−1)² also represents two consecutive integers, as does (2m)²+(2m+1)².

   Final answer: (i) s²+2; (ii) m²+(m+1)², m²+(m−1)² and (2m)²+(2m+1)²

4. Question 4. Why do the diagonal products in a 2×2 calendar block differ by 7?

   Step-by-step solution: Write the block as a,a+1 in the top row and a+7,a+8 in the bottom. Diagonal products: a(a+8) and (a+1)(a+7)=a²+8a+7.

   Final answer: The second diagonal product is always 7 greater.

5. Question 5. Decide which four statements about algebraic expressions are true.

   Step-by-step solution: (i) (k+1)(k+2)−(k+3)=k²+2k−1, not always 2: false. (ii) (2q+1)(2q−3) is odd for integer q, so not divisible by 4: false. (iii) (2n)²=4n²; (2n+1)²=4n(n+1)+1 and n(n+1) even, hence 1 mod 8: true. (iv) (6n+2)²−(4n+3)²=20n²+13? Expand explicitly: 36n²+24n+4−16n²−24n−9=20n²−5. This is 5 less than a square only if 20n² is square, which is not generally true: false.

   Final answer: (i) False; (ii) False; (iii) True; (iv) False

6. Question 6. Find remainders of the sum, difference and product of numbers leaving 3 and 5 modulo 7.

   Step-by-step solution: Let numbers be 7a+3 and 7b+5. Sum has residue 8≡1; first minus second has residue −2≡5; product has residue 15≡1 modulo 7.

   Final answer: Sum 1; difference 5 (first minus second); product 1

7. Question 7. Square the middle of three consecutive integers and subtract the product of the others.

   Step-by-step solution: Let integers n−1,n,n+1. Compute n²−(n−1)(n+1)=n²−(n²−1)=1.

   Final answer: Always 1

8. Question 8. Add two numbers, then multiply their sum by half that sum.

   Step-by-step solution: Let numbers a,b. Sum S=a+b; half sum=S/2; product S×S/2=S²/2.

   Final answer: (a+b)²/2

9. Question 9. Compare 14×26 with 16×24, and 25×75 with 26×74 without long multiplication.

   Step-by-step solution: (i) Around 20: 14×26=20²−6²=364; 16×24=20²−4²=384. (ii) 25×75=1875; 26×74=(25+1)(75−1)=1875+75−25−1=1924.

   Final answer: (i) 16×24 greater; (ii) 26×74 greater

10. Question 10. Write an expression for the tiled path around two square garden plots.

    Step-by-step solution: The result depends on the arrangement and width w shown in the official figure. Without fixing which sides touch and the outside dimensions, a unique area formula cannot be derived.

    Final answer: Diagram-dependent: use the original textbook figure before assigning a formula.

11. Question 11. Extend the two visual patterns and express Step 10 and Step y.

    Step-by-step solution: These are diagram-specific sequences. Count the basic units in Steps 1–3 from the original figure and test the proposed rule against all three. The diagrams are not reproduced in this guide.

    Final answer: Pending figure-by-figure verification; no invented Step 10 count.

Source and accuracy audit

Source: NCERT Ganita Prakash Grade 8 Part I Chapter 6, Reprint 2026–27. All three numbered Figure it Out sets (5, 4 and 11 questions) are covered, with the final two diagram-dependent questions explicitly marked for figure verification. The chapter also includes in-text Math Talk and extended Coin Conjoin puzzles; these require a separate full visual audit and are not claimed to be exhaustively solved here. Original exam practice is clearly distinguished from textbook questions.