AP 8th Maths Textbook Solutions – 6. We Distribute, Yet Things Multiply (2026-27)

6 WE DISTRIBUTE, YET THINGS MULTIPLY

6.1 Some Properties of Multiplication

• 1. By how much does the product increase if the first number (23) is increased by 1?
  The product increases by the value of the second number, which is 27. (Since 24 × 27 = 23 × 27 + 27).
• 2. What if the second number (27) is increased by 1?
  The product increases by the value of the first number, which is 23. (Since 23 × 28 = 23 × 27 + 23).
• 3. How about when both numbers are increased by 1?
  The product increases by the sum of the original two numbers plus 1. In this case, it increases by 23 + 27 + 1 = 51.
• 4. Do you see a pattern that could help generalise our observations to the product of any two numbers?
  Yes, for any two numbers a and b, if a is increased by 1, the product increases by b. If b is increased by 1, the product increases by a. If both are increased by 1, the product increases by a + b + 1.

• 1. What would we get if we had expanded (a + 1)(b + 1) by first taking (b + 1) as a single term? Try it?
  Taking (b + 1) as a single term:
  (a + 1)(b + 1) = a(b + 1) + 1(b + 1)
  = ab + a + b + 1
  This gives the exact same final result.

• 1. 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?
  Yes, there will be a change. The product (a + 1)(b - 1) expands to ab - a + b - 1. So, the product changes by an amount equal to (b - a - 1).
• 2. Will the product always increase? Find 3 examples where the product decreases.
  No, it will not always increase. If (b - a - 1) is a negative value, the product decreases.
  Example 1: 10 × 5 → 11 × 4 = 44 (decreased by 6).
  Example 2: 8 × 2 → 9 × 1 = 9 (decreased by 7).
  Example 3: 4 × 3 → 5 × 2 = 10 (decreased by 2).
• 3. What happens when a and b are negative integers? Check by substituting different values for a and b.
  The algebraic rule still holds true.
  For a = -4, b = -5:
  Original product: (-4) × (-5) = 20.
  New product: (-4 + 1)(-5 - 1) = (-3)(-6) = 18.
  Change = b - a - 1 = -5 - (-4) - 1 = -2.
  The original product 20 decreased by 2 to become 18.
• 4. By how much will the product of two numbers change if one of the numbers is increased by m and the other by n?
  The product (a + m)(b + n) expands to ab + an + bm + mn. The product increases by the amount (an + bm + mn).

• 1. This identity can be used to find how products change when the numbers being multiplied are increased or decreased by any amount. Can you see how this identity can be used when one or both numbers are decreased?
  We can substitute negative values for m and/or n into the identity (a + m)(b + n) = ab + mb + an + mn.

• 1. 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.
  Verify the answers by finding the products without converting the subtractions to additions.
  (i) Here m = -2, n = 3.
  Change = mb + an + mn = (-2)b + a(3) + (-2)(3) = 3a - 2b - 6.
  (ii) Here m = -3, n = -4.
  Change = mb + an + mn = (-3)b + a(-4) + (-3)(-4) = -4a - 3b + 12.
• 2. Expand (i) (a - u)(b + v) and (ii) (a - u)(b - v).
  (i) (a - u)(b + v) = ab + av - ub - uv
  (ii) (a - u)(b - v) = ab - av - ub + uv
• 3. Can any two terms be added to get a single term? For example, can &frac32;a² and &frac310;a be added to get a single term?
  No. Terms with different powers of a variable (like a² and a) are "unlike terms" and cannot be combined into a single term.

• 1. a × a² = a³ (why?)
  According to the laws of exponents, when multiplying terms with the same base, you add their exponents: a¹ × a² = a¹⁺² = a³.

Figure it Out - 6.1

• 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.
  Top Left: (p-1)(q-1)
  Top Middle: (p-1)q
  Top Right: (p-1)(q+1)
  Middle Left: p(q-1)
  Middle Right: p(q+1)
  Bottom Left: (p+1)(q-1)
  Bottom Middle: (p+1)q
  Bottom Right: (p+1)(q+1)

• 2. Expand the following products:
  (i) (3 + u)(v - 3)
  (ii) ⅔(15 + 6a)
  (iii) (10a + b)(10c + d)
  (iv) (3 - x)(x - 6)
  (v) (-5a + b)(c + d)
  (vi) (5 + z)(y + 9)
  (i) 3v - 9 + uv - 3u
  (ii) 10 + 4a
  (iii) 100ac + 10ad + 10bc + bd
  (iv) 3x - 18 - x² + 6x = -x² + 9x - 18
  (v) -5ac - 5ad + bc + bd
  (vi) 5y + 45 + zy + 9z
• 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.
  For the product to remain unchanged, (a+2)(b-4) = ab, which simplifies to b - 2a = 4.
  Example 1: a = 1, b = 6 → 1 × 6 = 3 × 2 = 6.
  Example 2: a = 2, b = 8 → 2 × 8 = 4 × 4 = 16.
  Example 3: a = 3, b = 10 → 3 × 10 = 5 × 6 = 30.
• 4. Expand (i) (a + ab - 3b²)(4 + b), and (ii) (4y + 7)(y + 11z - 3).
  (i) 4a + ab + 4ab + ab² - 12b² - 3b³ = 4a + 5ab + ab² - 12b² - 3b³
  (ii) 4y² + 44yz - 12y + 7y + 77z - 21 = 4y² + 44yz - 5y + 77z - 21
• 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?
  (i) a² - b²
  (ii) a³ - b³
  (iii) a&sup4; - b&sup4;
  The pattern reveals that multiplying (a - b) by the sum of decreasing powers of 'a' and increasing powers of 'b' gives aⁿ - bⁿ.
  The next identity would be: (a - b)(a&sup4; + a³b + a²b² + ab³ + b&sup4;) = a⁵ - b⁵.
• 6. Use the following multiplications to find the product of a number with 11 in a single step: (a) 3874 × 11 (b) 5678 × 11.
  (a) 3874 × 11 = 42614
  (b) 5678 × 11 = 62458

• 1. Describe a general rule to multiply a number (of any number of digits) by 11 and write the product in one line.
  Write down the last digit of the number. Working right to left, add each adjacent pair of digits, writing down the units digit of their sum and carrying over any tens. Finally, write the first digit (plus any carry from the previous step).
• 2. Evaluate (i) 94 × 11, (ii) 495 × 11, (iii) 3279 × 11, (iv) 4791256 × 11.
  (i) 1034
  (ii) 5445
  (iii) 36069
  (iv) 52703816
• 3. Can we come up with a similar rule for multiplying a number by 101? Multiply 3874 by 101 in one line.
  Yes. Instead of adding adjacent digits, you add digits that are separated by one position (skipping the immediate neighbor).
  3874 × 101 = 391274.

• 1. 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, ...
  To multiply by 101, add the number shifted by two decimal places to itself. To multiply by 1001, add the number shifted by three decimal places, and so on.
• 2. Use this to find (i) 89 × 101, (ii) 949 × 101, (iii) 265831 × 1001, (iv) 1111 × 1001, (v) 9734 × 99 and (vi) 23478 × 999.
  (i) 8989
  (ii) 95849
  (iii) 266096831
  (iv) 1112111
  (v) 9734 × (100 - 1) = 963666
  (vi) 23478 × (1000 - 1) = 23454522
• 3. 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?
  Yes, by adding the areas of the constituent parts: (60 + 5)² = 60² + 2(60 × 5) + 5² = 3600 + 600 + 25 = 4225 sq. units.

• 1. What if we write 65² as (30 + 35)² or (52 + 13)²? Draw the figures and check the area that you get.
  You will get the exact same area.
  (30 + 35)² = 30² + 2(30 × 35) + 35² = 900 + 2100 + 1225 = 4225.
  (52 + 13)² = 52² + 2(52 × 13) + 13² = 2704 + 1352 + 169 = 4225.
• 2. If a and b are any two integers, is (a + b)² always greater than a² + b²? If not, when is it greater?
  It is not always greater. (a + b)² is greater than a² + b² only when 2ab > 0 (i.e., when a and b have the same sign). If either is zero, they are equal. If they have opposite signs, (a + b)² is smaller.
• 3. Use Identity 1A to find the values of 104², 37².
  104² = (100 + 4)² = 100² + 2(100 × 4) + 4² = 10000 + 800 + 16 = 10816.
  37² = (30 + 7)² = 30² + 2(30 × 7) + 7² = 900 + 420 + 49 = 1369.
• 4. Use Identity 1A to write the expressions for the following: (i) (m + 3)² (ii) (6 + p)².
  (i) m² + 6m + 9
  (ii) 36 + 12p + p²
• 5. Expand (6x + 5)².
  (6x)² + 2(6x)(5) + 5² = 36x² + 60x + 25.

• 1. Expand (3j + 2k)² using both the identity and by applying the distributive property.
  Using Identity: (3j)² + 2(3j)(2k) + (2k)² = 9j² + 12jk + 4k².
  Using Distribution: (3j + 2k)(3j + 2k) = 9j² + 6jk + 6kj + 4k² = 9j² + 12jk + 4k².
• 2. Can we use 60² (=3600) and 5² (=25) to find the value of (60 - 5)² or 55²?
  Yes, using the identity (a - b)² = a² - 2ab + b².
  (60 - 5)² = 60² - 2(60 × 5) + 5² = 3600 - 600 + 25 = 3025.

• 1. We can also use the expansion of (a + b)² to find the expansion of (a - b)². Think how.
  By rewriting (a - b)² as (a + (-b))² and substituting (-b) for b in the identity: a² + 2(a)(-b) + (-b)² = a² - 2ab + b².
• 2. Find the general expansion of (a - b)² using geometry, as we did for 55².
  Start with a large square of side 'a' (Area = a²). Subtract two rectangles of dimensions a × b. Because the small corner square of b × b was subtracted twice in this process, we must add it back once to get the correct remaining area. Thus, the area is a² - 2ab + b².
• 3. Use the identity (a - b)² to find the values of (a) 99² and (b) 58².
  (a) 99² = (100 - 1)² = 10000 - 2(100) + 1 = 9801.
  (b) 58² = (60 - 2)² = 3600 - 2(120) + 4 = 3364.
• 4. Expand the following using both Identity 1B and by applying the distributive property: (i) (b - 6)² (ii) (-2a + 3)² (iii) (7y - ¾z)²
  (i) b² - 12b + 36
  (ii) 4a² - 12a + 9
  (iii) 49y² - &frac{21y}{2z} + &frac{9}{16z²}
• 5. Take a pair of natural numbers. Calculate the sum of their squares. Can you write twice this sum as a sum of two squares?
  Yes, this follows the pattern 2(a² + b²) = (a + b)² + (a - b)². For example, for 4 and 2: 2(16 + 4) = 40. And (4 + 2)² + (4 - 2)² = 36 + 4 = 40.
• 6. Here is a related pattern. Try to describe the pattern using algebra to determine if the pattern always holds.
  The pattern shows the difference of two squares. Using algebra: a² - b² = (a + b)(a - b). This identity holds true for all numbers.

• 1. Use Identity 1C to calculate 98 × 102, and 45 × 55.
  98 × 102 = (100 - 2)(100 + 2) = 100² - 2² = 10000 - 4 = 9996.
  45 × 55 = (50 - 5)(50 + 5) = 50² - 5² = 2500 - 25 = 2475.
• 2. Show that (a + b) × (a - b) = a² - b² geometrically.
  Take a rectangle with sides (a + b) and (a - b). The piece measuring b × (a - b) can be cut from the end and rotated to fit underneath the a × (a - b) section. This forms a large square of side 'a' that is missing a small corner square of side 'b', proving the area is a² - b².
• 3. a² = (a + b)(a - b) + b². Why is this identity true?
  Because the expansion of (a + b)(a - b) is a² - b². Adding b² to this gives a² - b² + b², which leaves exactly a².

Figure it Out - 6.2

• 1. Which is greater: (a - b)² or (b - a)²? Justify your answer.
  They are always exactly equal. Squaring a negative number yields a positive result, so (a - b)² = (-(b - a))² = (b - a)².
• 2. Express 100 as the difference of two squares.
  We need to find x and y such that x² - y² = 100, which means (x - y)(x + y) = 100.
  If we let (x - y) = 2 and (x + y) = 50, then x = 26 and y = 24.
  Therefore, 26² - 24² = 676 - 576 = 100.
• 3. Find 406², 72², 145², 1097², and 124² using the identities you have learnt so far.
  406² = (400 + 6)² = 160000 + 4800 + 36 = 164836.
  72² = (70 + 2)² = 4900 + 280 + 4 = 5184.
  145² = (140 + 5)² = 19600 + 1400 + 25 = 21025.
  1097² = (1100 - 3)² = 1210000 - 6600 + 9 = 1203409.
  124² = (120 + 4)² = 14400 + 960 + 16 = 15376.
• 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.
  Yes, these identities hold true for all real numbers, including negative integers and fractions, because the fundamental laws of algebra (distributive, commutative, associative) apply universally to all real numbers.

6.3 Mind the Mistake, Mend the Mistake

• 5. Check each of the simplifications and see if there is a mistake. If there is a mistake, try to explain what could have gone wrong. Then write the correct expression.
  1. Correct: 15p² - 6pq. Mistake: Did not multiply the terms properly.
  2. Correct: 2x - 2 + 3x + 12 = 5x + 10. Mistake: Forgot to distribute the constants to the second terms inside the brackets.
  3. Correct: y + 2y + 4 = 3y + 4. Mistake: Incorrectly assumed y + 2(y + 2) is (y + 2)(y + 2).
  4. Correct: 25m² + 60mn + 36n². Mistake: Missed the middle 2ab term.
  5. Correct as written. (-q)² + 2(-q)(2) + 2² = q² - 4q + 4. No mistake.
  6. Correct: 18abc. Mistake: Distributed over multiplication instead of addition.
  7. Correct as written. No mistake.
  8. Correct: 5w² + 6w. Mistake: Incorrectly added unlike terms.
  9. Correct: 5a³ + 6a²b + 6ab². Mistake: Incorrectly added unlike terms 6a²b and 6ab².
  10. Correct as written. No mistake.
  11. Correct: ab + 4a + 2b + 8. Mistake: Failed to fully foil/distribute all terms.
  12. Correct: a(h² + ab + ab²). Mistake: Factored out an incorrect common term.

• 1. Observe the pattern in the figure below. Draw the next figure in the sequence. How many circles does it have? How many total circles are there in Step 10? Write an expression for the number of circles in Step k.
  The next figure (Step 4) is a 5 × 4 rectangle minus 1 corner circle, giving 19 circles (or using the formula: 4² + 2(4) = 24).
  Step 10 will have (10 + 1)² - 1 = 121 - 1 = 120 circles.
  The expression for Step k is (k + 1)² - 1, which simplifies to k² + 2k.

• 1. Use this formula to find the number of circles in Step 15.
  k² + 2k = 15² + 2(15) = 225 + 30 = 255 circles.
• 2. How many square tiles are there in each figure?
  Step 1 has 8 tiles.
  Step 2 has 12 tiles.
  Step 3 has 16 tiles.
• 3. How many are there in Step 4 of the sequence? What about Step 10?
  Step 4 will have a 6 × 6 outer square minus a 4 × 4 inner square = 36 - 16 = 20 tiles.
  Step 10 will have a 12 × 12 outer square minus a 10 × 10 inner square = 144 - 100 = 44 tiles.

• 1. Write an algebraic expression for the number of tiles in Step n. Share your methods with the class.
  Using outer square minus inner square method: (n + 2)² - n² = n² + 4n + 4 - n² = 4n + 4.
• 2. Find the area of the (interior) shaded region in the figure below. All four rectangles have same dimensions.
  The total region is a square of side (m + n), with area (m + n)². Subtracting the 4 rectangles (each of area mn) gives: (m + n)² - 4mn.
• 3. By expanding both expressions, check that (m + n)² - 4mn = (n - m)².
  (m + n)² - 4mn = m² + 2mn + n² - 4mn
  = m² - 2mn + n²
  = (n - m)². The expressions are equivalent.
• 4. Find out the area of the region with slanting lines in the figure. All three rectangles have the same dimensions (Fig. 1).
  Anusha's method: Area of ABCD = x². Area of EFGH = xy. Required area = x² - xy.

• 1. By expanding the expressions, verify that all three expressions are equivalent. If x = 8 and y = 3 find the area of the shaded region.
  Expression 1: x² - xy.
  Expression 2: x(x + 2y) - 3xy = x² + 2xy - 3xy = x² - xy.
  Expression 3: 2[x(x - y)/2] = x(x - y) = x² - xy.
  All three are perfectly equivalent. If x = 8 and y = 3, Area = 8² - (8)(3) = 64 - 24 = 40.
• 2. 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.
  The dashed region forms a rectangle with width (s - r) and height (p - r).
  Area = (s - r)(p - r) = sp - sr - pr + r².
  Substituting: Area = (9 - 3.5)(6 - 3.5) = 5.5 × 2.5 = 13.75.

Figure it Out - 6.3

• 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)
  (i) (40 + 6)² = 1600 + 480 + 36 = 2116.
  (ii) (400 - 3)(400 + 3) = 160000 - 9 = 159991.
  (iii) (100 - 9)² = 10000 - 1800 + 81 = 8281.
  (iv) (44 - 1)(44 + 1) = 1936 - 1 = 1935.
• 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 - ½)²
  (vi) (7p) × (3r) × (p + 2)
  (i) p² + 10p - 11
  (ii) 9a² - 81b²
  (iii) -6y² - 23y - 20
  (iv) 36x² + 60xy + 25y²
  (v) 4x² - 2x + ¼
  (vi) 21p²r + 42pr
• 3. For each statement identify the appropriate algebraic expression(s).
  (i) Two more than a square number.
  (ii) The sum of the squares of two consecutive numbers
  (i) s² + 2
  (ii) m² + (m + 1)²
• 4. Consider any 2 by 2 square of numbers in a calendar. Find products of numbers lying along each diagonal. Do this for the other 2 by 2 squares. What do you observe about the diagonal products? Explain why this happens.
  The difference between the two diagonal products in any 2x2 calendar square is always 7.
  Explanation: Let the top-left number be 'a'. The others are (a+1), (a+7), and (a+8).
  Diagonal 1: a(a+8) = a² + 8a.
  Diagonal 2: (a+1)(a+7) = a² + 8a + 7.
  The difference is clearly exactly 7 regardless of 'a'.

• 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.
  (i) False. It expands to k² + 2k - 1, which changes depending on k.
  (ii) False. It expands to 4q² + 8q + 3, which is 3 more than a multiple of 4.
  (iii) True. (2n)² = 4n². And (2n + 1)² = 4n(n + 1) + 1. Since n(n+1) is always even, 4 times an even number is a multiple of 8, making it 8k + 1.
  (iv) False. The expansion is 20n² - 5, which is not always 5 less than a perfect square.
• 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?
  Sum remainder: (3 + 5) = 8. 8 divided by 7 leaves a remainder of 1.
  Difference remainder: (5 - 3) = 2 (or -2, which translates to a remainder of 5).
  Product remainder: (3 × 5) = 15. 15 divided by 7 leaves a remainder of 1.
• 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.
  The result is always exactly 1.
  Algebraically: Let the consecutive numbers be n-1, n, n+1.
  Equation: n² - (n - 1)(n + 1) = 1.
  Expansion: n² - (n² - 1) = 1, which simplifies to 1 = 1.
• 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.
  Expression: (a + b) × &frac{a + b}{2}
  Proof: Multiplying the numerators gives (a + b)². Therefore, it results in &frac{(a + b)²}{2}, which is precisely half of the square of the sum.
• 9. Which is larger? Find out without fully computing the product.
  (i) 14 × 26 or 16 × 24
  (ii) 25 × 75 or 26 × 74
  (i) 16 × 24 is larger. Both center around 20. 14 × 26 = 20² - 6² = 400 - 36. 16 × 24 = 20² - 4² = 400 - 16. Subtracting a smaller square yields a larger number.
  (ii) 26 × 74 is larger. They center around 50. 50² - 25² vs 50² - 24².
• 10. A tiny park is coming up in Dhauli. The plan is shown in the figure. Write an expression for the area that needs to be tiled.
  The total width is (2g + 3w) and the total height is (g + 2w).
  Total area = (2g + 3w)(g + 2w). Area of green squares = 2g².
  Tiled area = (2g + 3w)(g + 2w) - 2g² = 2g² + 7gw + 6w² - 2g² = 7gw + 6w².

• 11. For each pattern shown below,
  (i) Draw the next figure in the sequence.
  (ii) How many basic units are there in Step 10?
  (iii) Write an expression to describe the number of basic units in Step y.
  Left Pattern (Plus Shapes):
  (ii) Step 10 has 4(10) + 1 = 41 basic units.
  (iii) Step y has 4y + 1 basic units.
  
  Right Pattern (Blue Squares):
  (ii) The sequence is 5, 10, 17, which matches (n+1)² + 1. Step 10 has (10+1)² + 1 = 121 + 1 = 122 basic units.
  (iii) Step y has (y + 1)² + 1 basic units.

Puzzle Time! Coin Conjoin

• 1. The task is to turn the triangle upside down by moving one coin at a time. How many moves are needed? What is the minimum number of moves? Find out the minimum possible moves needed to flip the next bigger triangle having 15 coins. Try the same for bigger triangular numbers. Is there a simple way to calculate the minimum number of coin moves needed for any such triangular arrangement?
  For a 10-coin triangle, the minimum is 3 moves (moving the 3 corner coins).
  For a 15-coin triangle, the minimum is 5 moves.
  A simple rule to calculate the minimum moves for any standard coin triangle is to divide the total number of coins by 3, and discard any decimal remainder (integer division).
  Example: 10 / 3 = 3.33 (3 moves). 15 / 3 = 5 (5 moves).

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