Questions: We Distribute, Yet Things Multiply
We Distribute, Yet Things Multiply24 min readLesson 6 of 6
Q1 State the distributive property of multiplication over addition and use it to find 23 × 28.
Answer (English): The distributive property says a(b + c) = ab + ac for any numbers a, b and c. Write 28 as 27 + 1. Then 23 × 28 = 23(27 + 1) = 23 × 27 + 23 = 621 + 23 = 644. The same expansion shows why increasing 27 by 1 adds exactly one copy of 23 to the product.
Hinglish: Distributive property ka matlab a(b + c) = ab + ac hai. 28 ko 27 + 1 likho, to 23 × 28 = 23 × 27 + 23 = 621 + 23 = 644. Isi wajah se 27 ko 1 badhane par product mein ek 23 judta hai.
Q2 Expand (a + 1)(b + 1) and explain what the extra terms mean.
Answer (English): Using distributivity twice, (a + 1)(b + 1) = ab + a + b + 1. Compared with ab, the product has grown by a + b + 1. In the rectangle picture, a and b are the two added strips along the sides and the 1 is the small square at the corner where the strips overlap.
Hinglish: Distributivity do baar lagao: (a + 1)(b + 1) = ab + a + b + 1. Product ab se a + b + 1 zyada ho gaya. Rectangle mein a aur b do side strips hain aur 1 woh chhota corner square hai jahan strips milti hain.
Q3 State Identity 1 and check it for a = 5, b = 7, m = 2, n = −3.
Answer (English): Identity 1 expands the product of two changed factors into four terms. Substituting gives (5 + 2)(7 + (−3)) = 7 × 4 = 28. The right side is 35 + 14 − 15 − 6 = 28. Both sides agree, so the identity handles a negative n correctly.
Hinglish: Identity 1 do changed factors ke chaar terms deti hai. Values daalo: (5 + 2)(7 − 3) = 7 × 4 = 28. Right side = 35 + 14 − 15 − 6 = 28. Dono barabar hain, matlab negative n ke saath bhi identity chalti hai.
Q4 Expand (3x − 2y)².
Answer (English): Use Identity 1B: (a − b)² = a² − 2ab + b² with a = 3x and b = 2y. Then (3x)² − 2(3x)(2y) + (2y)² = 9x² − 12xy + 4y². A common error is to leave out the middle term and write 9x² + 4y², which is only true when xy = 0.
Hinglish: Identity 1B lagao: (a − b)² = a² − 2ab + b², yahan a = 3x aur b = 2y. To 9x² − 12xy + 4y². Middle term bhool kar 9x² + 4y² likhna galti hai; woh sirf xy = 0 par sach hota hai.
Q5 Expand (2a + 5)(2a − 5) using Identity 1C.
Answer (English): Identity 1C says (a + b)(a − b) = a² − b². With 2a in place of a and 5 in place of b, the product is (2a)² − 5² = 4a² − 25. The two middle terms cancel, so no middle term appears in the answer.
Hinglish: Identity 1C: (a + b)(a − b) = a² − b². Yahan a ki jagah 2a aur b ki jagah 5 hai, isliye (2a)² − 25 = 4a² − 25. Beech ke dono terms cancel ho jate hain, isliye middle term nahi aata.
Q6 What are like terms? Simplify 4a²b + 3a²b − 2ab + 7.
Answer (English): Like terms have exactly the same letter part, even if their numerical coefficients differ. 4a²b and 3a²b are like terms and add to 7a²b. The term −2ab has a different letter part and the constant 7 has none, so neither can join. The simplified expression is 7a²b − 2ab + 7.
Hinglish: Like terms woh hote hain jinka letter part bilkul same ho, chahe numbers alag hon. 4a²b aur 3a²b like terms hain aur 7a²b bante hain. −2ab ka letter part alag hai aur 7 constant hai, isliye unhe jod nahi sakte. Simplified answer: 7a²b − 2ab + 7.
Q7 Explain Pattern 1: 2(a² + b²) = (a + b)² + (a − b)².
Answer (English): Expand both squares with Identities 1A and 1B: (a + b)² = a² + 2ab + b² and (a − b)² = a² − 2ab + b². Adding them gives a² + a² + 2ab − 2ab + b² + b² = 2a² + 2b². The middle terms cancel, which is why the identity holds for every a and b.
Hinglish: Dono squares ko Identity 1A aur 1B se expand karo: (a + b)² = a² + 2ab + b² aur (a − b)² = a² − 2ab + b². Jodne par 2a² + 2b² milta hai kyunki 2ab aur −2ab cancel ho jate hain. Isliye yeh identity har a aur b ke liye sach hai.
Q8 Use distributivity to compute 47 × 11, 68 × 101 and 24 × 99.
Answer (English): For 11 = 10 + 1: 47 × 10 + 47 = 470 + 47 = 517. For 101 = 100 + 1: 68 × 100 + 68 = 6800 + 68 = 6868. For 99 = 100 − 1: 24 × 100 − 24 = 2400 − 24 = 2376. The same trick works for 1001 and 999 by using 1000 ± 1.
Hinglish: 11 = 10 + 1 se 47 × 11 = 470 + 47 = 517. 101 = 100 + 1 se 68 × 101 = 6800 + 68 = 6868. 99 = 100 − 1 se 24 × 99 = 2400 − 24 = 2376. Isi trick se 1001 aur 999 bhi 1000 ± 1 karke nikalte hain.
Q9 Use Sridharacharya's form a² = (a + b)(a − b) + b² to find 52².
Answer (English): Choose b = 2 so that a + b = 54 and a − b = 50 are round numbers. Then 52² = (52 + 2)(52 − 2) + 2² = 54 × 50 + 4 = 2700 + 4 = 2704. Choosing b = 3 would give 55 × 49 + 9 = 2695 + 9 = 2704 as well, which is a useful check.
Hinglish: b = 2 chuno taaki a + b = 54 aur a − b = 50 round numbers banein. Phir 52² = 54 × 50 + 4 = 2700 + 4 = 2704. b = 3 se 55 × 49 + 9 = 2695 + 9 = 2704 aata hai, jo check ke liye achha hai.
Q10 A pattern has k² + 2k circles in Step k. Why do four different expressions in the chapter all give this count?
Answer (English): The chapter lists (k + 1)² − 1, k² + 2k, k(k + 1) + k and k(k + 2). Expanding each gives the same expression: (k + 1)² − 1 = k² + 2k, k(k + 1) + k = k² + k + k = k² + 2k, and k(k + 2) = k² + 2k. The four methods count the same pattern in different ways, so they must agree.
Hinglish: Chapter mein chaar expressions hain: (k + 1)² − 1, k² + 2k, k(k + 1) + k aur k(k + 2). Sabko expand karo to same k² + 2k milta hai. Chaaron methods ek hi pattern ko alag tarah se gin rahe hain, isliye result same hona chahiye.
Quiz pointer
The five-question quiz for this chapter is separate from these answers. Take it without looking back, then redo any question you missed with the identity box in front of you. For each wrong option, write the correct expansion by hand, because that is where the mistake usually becomes visible.