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.

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…

By Syllab.in · Updated Jun 14, 2026

Key Questions Covered:

Expand: 3(2x + 5)
Simplify: (x + 2)(x + 3)
Evaluate the identity: (a + b)² = a² + 2ab + b² for a = 2, b = 3
Multiply: (2x + 3)(x - 2)
Use the identity (a - b)² = a² - 2ab + b² to find (5 - 3)²
Expand: -2(3x - 4y)
+ 2 more questions in the full chapter

Solutions Summary:

Question	Status
Expand: 3(2x + 5)	✓ Solved
Simplify: (x + 2)(x + 3)	✓ Solved
Evaluate the identity: (a + b)² = a² + 2ab + b² for a = 2…	✓ Solved
Multiply: (2x + 3)(x - 2)	✓ Solved
Use the identity (a - b)² = a² - 2ab + b² to find (5 - 3)²	✓ Solved
Expand: -2(3x - 4y)	✓ Solved

Showing 6 of 8 questions

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 + ...

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

Showing 6 of 8 questions. Visit the full page for complete solutions.