We Distribute Yet Things Multiply — Class 8 Mathematics NCERT Solutions (Free)

Free step-by-step NCERT solutions for Class 8 Mathematics chapter "We Distribute Yet Things Multiply" — 8 important questions with detailed answers for CBSE board exam preparation.

We Distribute Yet Things Multiply — Class 8 Mathematics NCERT Solutions (Free)

Free step-by-step NCERT solutions for Class 8 Mathematics chapter "We Distribute Yet Things Multiply" — 8 important questions with detailed answers for CBSE board exam preparation.

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TL;DR: Free step-by-step NCERT solutions for Class 8 Mathematics chapter "We Distribute Yet Things Multiply" — 8 important questions with detailed answers fo…

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NCERT Solutions for Class 8 Mathematics — We Distribute Yet Things Multiply. Step-by-step answers to all 8 textbook questions from this chapter, written for CBSE board preparation and free to use.

We Distribute Yet Things Multiply — All 8 Questions Solved

Q1. Expand: 3(2x + 5)

We use the distributive property to expand this expression.

Step 1: Recall the distributive property
a(b + c) = ab + ac

Step 2: Identify a, b, and c
a = 3
b = 2x
c = 5

Step 3: Apply the distributive property
3(2x + 5) = 3(2x) + 3(5)

Step 4: Multiply
3(2x) = 6x
3(5) = 15

Step 5: Combine
3(2x + 5) = 6x + 15

Answer: 3(2x + 5) = 6x + 15

Q2. Simplify: (x + 2)(x + 3)

We need to multiply two binomials using the distributive property (FOIL method).

Step 1: Use the distributive property twice
(x + 2)(x + 3) = x(x + 3) + 2(x + 3)

Step 2: Distribute x
x(x + 3) = x² + 3x

Step 3: Distribute 2
2(x + 3) = 2x + 6

Step 4: Combine like terms
(x + 2)(x + 3) = x² + 3x + 2x + 6 = x² + 5x + 6

Step 5: Alternative - FOIL method
First: x × x = x²
Outer: x × 3 = 3x
Inner: 2 × x = 2x
Last: 2 × 3 = 6
Result: x² + 3x + 2x + 6 = x² + 5x + 6

Answer: (x + 2)(x + 3) = x² + 5x + 6

Q3. Evaluate the identity: (a + b)² = a² + 2ab + b² for a = 2, b = 3

We verify the algebraic identity by substituting values.

Step 1: Write the identity
(a + b)² = a² + 2ab + b²

Step 2: Substitute a = 2 and b = 3
Left side: (2 + 3)² = (5)² = 25

Step 3: Evaluate the right side
a² + 2ab + b² = 2² + 2(2)(3) + 3²
= 4 + 2(6) + 9
= 4 + 12 + 9
= 25

Step 4: Compare
Left side = 25
Right side = 25
✓ Both sides are equal

Answer: (a + b)² = a² + 2ab + b² is verified. For a = 2, b = 3: both sides equal 25

Q4. Multiply: (2x + 3)(x - 2)

We multiply two binomials using the distributive property.

Step 1: Use the distributive property
(2x + 3)(x - 2) = 2x(x - 2) + 3(x - 2)

Step 2: Distribute 2x
2x(x - 2) = 2x² - 4x

Step 3: Distribute 3
3(x - 2) = 3x - 6

Step 4: Combine like terms
(2x + 3)(x - 2) = 2x² - 4x + 3x - 6
= 2x² - x - 6

Step 5: Verify using FOIL
First: 2x × x = 2x²
Outer: 2x × (-2) = -4x
Inner: 3 × x = 3x
Last: 3 × (-2) = -6
Result: 2x² - 4x + 3x - 6 = 2x² - x - 6 ✓

Answer: (2x + 3)(x - 2) = 2x² - x - 6

Q5. Use the identity (a - b)² = a² - 2ab + b² to find (5 - 3)²

We apply the algebraic identity (a - b)² = a² - 2ab + b².

Step 1: Identify a and b
a = 5
b = 3

Step 2: Apply the identity
(a - b)² = a² - 2ab + b²
(5 - 3)² = 5² - 2(5)(3) + 3²

Step 3: Calculate each term
5² = 25
2(5)(3) = 30
3² = 9

Step 4: Substitute
(5 - 3)² = 25 - 30 + 9 = 4

Step 5: Verify by direct calculation
(5 - 3)² = (2)² = 4 ✓

Answer: Using the identity (a - b)² = a² - 2ab + b², we get (5 - 3)² = 25 - 30 + 9 = 4

Q6. Expand: -2(3x - 4y)

We use the distributive property with a negative coefficient.

Step 1: Apply the distributive property
-2(3x - 4y) = -2(3x) - (-2)(4y)
= -2(3x) + 2(4y)

Step 2: Multiply
-2(3x) = -6x
2(4y) = 8y

Step 3: Combine
-2(3x - 4y) = -6x + 8y

Step 4: Verify by substituting a value, say x = 1, y = 1
Left side: -2(3(1) - 4(1)) = -2(3 - 4) = -2(-1) = 2
Right side: -6(1) + 8(1) = -6 + 8 = 2 ✓

Answer: -2(3x - 4y) = -6x + 8y

Q7. Simplify: (m + 4)(m - 4)

We multiply two binomials that follow the difference of squares pattern.

Step 1: Recognize the pattern
(m + 4)(m - 4) has the form (a + b)(a - b) where a = m and b = 4

Step 2: Use the difference of squares identity
(a + b)(a - b) = a² - b²

Step 3: Apply the identity
(m + 4)(m - 4) = m² - 4² = m² - 16

Step 4: Verify using FOIL
First: m × m = m²
Outer: m × (-4) = -4m
Inner: 4 × m = 4m
Last: 4 × (-4) = -16
Result: m² - 4m + 4m - 16 = m² - 16 ✓

Answer: (m + 4)(m - 4) = m² - 16

Q8. Simplify: 5x(2x + 3) - 2x(3x + 4)

We expand and simplify algebraic expressions.

Step 1: Expand the first term using the distributive property
5x(2x + 3) = 5x(2x) + 5x(3) = 10x² + 15x

Step 2: Expand the second term
2x(3x + 4) = 2x(3x) + 2x(4) = 6x² + 8x

Step 3: Write the subtraction
5x(2x + 3) - 2x(3x + 4) = 10x² + 15x - (6x² + 8x)

Step 4: Distribute the negative sign
= 10x² + 15x - 6x² - 8x

Step 5: Combine like terms
Like terms in x²: 10x² - 6x² = 4x²
Like terms in x: 15x - 8x = 7x

Step 6: Final answer
5x(2x + 3) - 2x(3x + 4) = 4x² + 7x

Answer: 5x(2x + 3) - 2x(3x + 4) = 4x² + 7x

We Distribute Yet Things Multiply — Practice MCQs with Answers

Attempt these 20 multiple-choice questions after working through the solutions above. Each carries the correct option and the reasoning behind it, so a wrong answer tells you which idea to revisit.

MCQ 1. What is the value of 7 × (10 + 2) using the distributive property?

  • A. 70
  • B. 84
  • C. 120
  • D. 72

Answer: B. 84 — The distributive property states a × (b + c) = a × b + a × c. So, 7 × (10 + 2) = 7 × 10 + 7 × 2 = 70 + 14 = 84.

MCQ 2. Simplify the expression 3(x + 5).

  • A. 3x + 5
  • B. 3x + 15
  • C. x + 15
  • D. 8x

Answer: B. 3x + 15 — According to the distributive property, multiply 3 by each term inside the parenthesis: 3 × x + 3 × 5 = 3x + 15.

MCQ 3. Expand the expression 4(2y - 3).

  • A. 8y - 3
  • B. 8y - 12
  • C. 2y - 12
  • D. 4y - 12

Answer: B. 8y - 12 — Apply the distributive property: 4 × 2y - 4 × 3 = 8y - 12.

MCQ 4. Multiply -5a by (a + 2b).

  • A. -5a^2 + 10ab
  • B. 5a^2 + 10ab
  • C. -5a^2 - 10ab
  • D. -5a + 10ab

Answer: C. -5a^2 - 10ab — Distribute -5a to both terms inside the parenthesis: (-5a) × a + (-5a) × 2b = -5a^2 - 10ab.

MCQ 5. Which property is demonstrated by a × (b + c) = a × b + a × c?

  • A. Commutative Property
  • B. Associative Property
  • C. Distributive Property
  • D. Identity Property

Answer: C. Distributive Property — This equation is the fundamental definition of the distributive property of multiplication over addition.

MCQ 6. Find the product of (x + 3) and (x + 5).

  • A. x^2 + 15
  • B. x^2 + 8x + 15
  • C. x^2 + 8x + 8
  • D. x^2 + 15x

Answer: B. x^2 + 8x + 15 — Multiply each term of the first binomial by each term of the second: x(x+5) + 3(x+5) = x^2 + 5x + 3x + 15 = x^2 + 8x + 15.

MCQ 7. Expand (2a - 1)(a + 4).

  • A. 2a^2 - 4
  • B. 2a^2 + 7a - 4
  • C. 2a^2 - 9a - 4
  • D. 2a^2 + 9a - 4

Answer: B. 2a^2 + 7a - 4 — Using the distributive property, multiply 2a by (a+4) and -1 by (a+4): 2a(a+4) - 1(a+4) = 2a^2 + 8a - a - 4 = 2a^2 + 7a - 4.

MCQ 8. What is the expansion of (y + 4)^2?

  • A. y^2 + 16
  • B. y^2 + 8y + 16
  • C. y^2 + 4y + 16
  • D. y^2 + 8y + 8

Answer: B. y^2 + 8y + 16 — Use the identity (a+b)^2 = a^2 + 2ab + b^2, where a=y and b=4. So, y^2 + 2(y)(4) + 4^2 = y^2 + 8y + 16.

MCQ 9. Expand (3x - 2)^2.

  • A. 9x^2 - 4
  • B. 9x^2 - 12x + 4
  • C. 9x^2 + 12x + 4
  • D. 3x^2 - 12x + 4

Answer: B. 9x^2 - 12x + 4 — Use the identity (a-b)^2 = a^2 - 2ab + b^2, where a=3x and b=2. Thus, (3x)^2 - 2(3x)(2) + 2^2 = 9x^2 - 12x + 4.

MCQ 10. Simplify (5p + q)(5p - q).

  • A. 25p^2 - q^2
  • B. 25p^2 + q^2
  • C. 5p^2 - q^2
  • D. 25p^2 - 10pq + q^2

Answer: A. 25p^2 - q^2 — This is in the form of the identity (a+b)(a-b) = a^2 - b^2. Here, a=5p and b=q, so the product is (5p)^2 - q^2 = 25p^2 - q^2.

MCQ 11. Calculate 98 × 15 using the distributive property.

  • A. 1470
  • B. 1500
  • C. 1485
  • D. 1370

Answer: A. 1470 — Rewrite 98 as (100 - 2). Then, apply the distributive property: (100 - 2) × 15 = 100 × 15 - 2 × 15 = 1500 - 30 = 1470.

MCQ 12. Simplify 2x(x + 3) - 5(x - 2).

  • A. 2x^2 + x - 10
  • B. 2x^2 + x + 10
  • C. 2x^2 - 10
  • D. 2x^2 + 11x - 10

Answer: B. 2x^2 + x + 10 — Distribute 2x into the first parenthesis and -5 into the second: (2x^2 + 6x) - (5x - 10) = 2x^2 + 6x - 5x + 10 = 2x^2 + x + 10.

MCQ 13. A rectangular field has a length of (3x + 2) meters and a width of (x + 1) meters. What is its area?

  • A. 3x^2 + 2
  • B. 3x^2 + 5x + 2
  • C. 3x^2 + 3x + 2
  • D. 3x^2 + 2x + 2

Answer: B. 3x^2 + 5x + 2 — Area of a rectangle = length × width. So, Area = (3x + 2)(x + 1) = 3x^2 + 3x + 2x + 2 = 3x^2 + 5x + 2 square meters.

MCQ 14. Evaluate (10.5)^2 using algebraic identities.

  • A. 100.25
  • B. 110.25
  • C. 105.00
  • D. 110.50

Answer: B. 110.25 — Write 10.5 as (10 + 0.5). Using (a+b)^2 = a^2 + 2ab + b^2, we get (10)^2 + 2(10)(0.5) + (0.5)^2 = 100 + 10 + 0.25 = 110.25.

MCQ 15. Find the product of 4xy and 5x.

  • A. 20xy
  • B. 20x^2y
  • C. 9x^2y
  • D. 20x^2

Answer: B. 20x^2y — Multiply the coefficients and then multiply the variables: (4 × 5) × (x × x) × y = 20x^2y.

MCQ 16. Multiply -2p by (p^2 - 3p + 7).

  • A. -2p^3 - 6p^2 + 14p
  • B. -2p^3 + 6p^2 - 14p
  • C. -2p^3 + 6p^2 + 14p
  • D. 2p^3 - 6p^2 + 14p

Answer: B. -2p^3 + 6p^2 - 14p — Distribute -2p to each term inside the parenthesis: -2p(p^2) - 2p(-3p) - 2p(7) = -2p^3 + 6p^2 - 14p.

MCQ 17. If (x + 5)^2 = x^2 + _ + 25, what is the missing term?

  • A. 5x
  • B. 10x
  • C. 25x
  • D. 5

Answer: B. 10x — Using the identity (a+b)^2 = a^2 + 2ab + b^2, where a=x and b=5, the middle term is 2ab = 2 × x × 5 = 10x.

MCQ 18. Calculate 103 × 104 using the identity (x+a)(x+b) = x^2 + (a+b)x + ab.

  • A. 10712
  • B. 10700
  • C. 11200
  • D. 10612

Answer: A. 10712 — Apply the identity (x+a)(x+b) with x=100, a=3, b=4. So, 100^2 + (3+4)×100 + 3×4 = 10000 + 700 + 12 = 10712.

MCQ 19. If a student buys 5 notebooks, each costing ₹x, and 5 pens, each costing ₹y, what is the total cost?

  • A. 5x + y
  • B. x + 5y
  • C. 5(x + y)
  • D. 5xy

Answer: C. 5(x + y) — The cost of 5 notebooks is 5x, and the cost of 5 pens is 5y. The total cost is 5x + 5y, which can be written as 5(x + y) using the distributive property.

MCQ 20. What is the product of (x - 1)(x + 1)(x^2 + 1)?

  • A. x^4 - 1
  • B. x^4 + 1
  • C. x^2 - 1
  • D. x^3 - 1

Answer: A. x^4 - 1 — First, multiply (x - 1)(x + 1) using the identity (a-b)(a+b) = a^2 - b^2, which gives x^2 - 1. Then, multiply (x^2 - 1)(x^2 + 1) again using the same identity, resulting in (x^2)^2 - 1^2 = x^4 - 1.

How to Use These Solutions

Attempt each question yourself first and only then compare with the worked answer. Marks in the CBSE board exam are awarded for the METHOD as much as the final result, so reproduce the steps rather than memorising the last line. Where a solution states a law, a formula or a definition, learn that wording — examiners look for it.

Related practice for this chapter: chapter MCQs, previous-year questions, revision notes and sample papers.

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