Exercise 11.3 — Binomial x Binomial
Multiplying a binomial by a binomial or trinomial.
What is Exercise 11.3 About?
Exercise 11.3 of Class 8 Mathematics (CBSE, Telangana & Andhra Pradesh syllabus) covers the multiplication of a binomial by another binomial or trinomial. This is a major step up from Exercise 11.2, where a monomial was multiplied by a polynomial. Here, both expressions being multiplied can have two or more terms.
The key method remains the distributive law, but now it is applied twice: every term in the first bracket is multiplied by every term in the second bracket. The total number of products before simplification equals the product of the number of terms in each bracket — so a binomial × binomial gives 4 raw products, and a binomial × trinomial gives 6.
(a + b)(c + d) = a×c + a×d + b×c + b×d ← Every term × every term
How Double Distribution Works — Textbook Introduction
The textbook opens Exercise 11.3 with two demonstration examples before the exercise questions. Understanding these fully makes all exercise problems straightforward.
Find the product of (2a + 3b) and (a − 3b)
Each term of the first bracket — 2a and 3b — is distributed across the entire second bracket.
Find the product of (3x − 5y) and (2x² − 3xy + 3y²)
Each of the two terms in the first bracket is distributed across all three terms of the trinomial, giving 2 × 3 = 6 raw products.
Like terms here: −9x²y and −10x²y → −19x²y. And 9xy² and 15xy² → 24xy².
Understanding the Method — Grid / FOIL Approach
For a binomial × binomial, many students find the term-by-term grid method helpful. For (A + B)(C + D), you can think of four cells:
| × | C (first term, 2nd bracket) | D (second term, 2nd bracket) |
|---|---|---|
| A (first term, 1st bracket) | A × C | A × D |
| B (second term, 1st bracket) | B × C | B × D |
Fill the four cells, then add all four products together, combining any like terms. This grid ensures you never miss a multiplication pair.
Question 1 — Multiply the Binomials (Parts i to iv)
All four parts in Question 1 follow the same pattern: distribute each term of the first binomial across the second binomial, multiply term by term, and collect like terms.
Here the terms of the first bracket are themselves two-variable products — kl and lm. The distribution method is identical.
The terms in the first bracket have squared variables. The same distribution method applies — multiply exponents by adding them.
Question 2 — Find the Product (Binomial × Trinomial)
Question 2 introduces binomial × trinomial multiplication. The first bracket has 2 terms and the second has 3, so distributing gives 2 × 3 = 6 raw products before collecting like terms.
Here the first bracket is a trinomial and the second is a binomial — the order is reversed from Q2(i) but the method is identical. Distribute each of the 3 terms in the first bracket across both terms of the second. This gives 3 × 2 = 6 raw products.
A trinomial × binomial with all multi-variable terms. Distribute all 3 terms of the first bracket across both terms of the second bracket.
The first bracket contains cubed variables. When multiplying by the terms of the trinomial, remember to add exponents for matching variables.
Question 3 — Simplify the Following Expressions
Question 3 tests the complete process: expand multiple products and then combine all results together, carefully collecting like terms from across the combined expression. These are the most involved problems in Exercise 11.3.
This requires expanding three products separately and then adding/subtracting them.
Expand each product
Product 2: (x + y)(x − 3y) = x² − 3xy + xy − 3y² = x² − 2xy − 3y²
Product 3: (y − 3x)(4x − 5y) = 4xy − 5y² − 12x² + 15xy = −12x² + 19xy − 5y²
Combine: P1 + P2 − P3
A binomial × trinomial with a very elegant simplification. This identity is one of the most important in algebra.
The most complex simplification in this exercise — three products to expand and combine.
Expand each product
Product 2: (a − b − c)(2a + 3c) = 2a² + 3ac − 2ab − 3bc − 2ac − 3c² = 2a² − 2ab + ac − 3bc − 3c²
Product 3: (6a + b)(2c − 3a − 5b) = 12ac − 18a² − 30ab + 2bc − 3ab − 5b² = −18a² − 33ab + 12ac + 2bc − 5b²
Combine: P1 − P2 + P3
Product 2: pq(p + q − r) = p²q + pq² − pqr
Total = p²q² − q²r² + 2pqr² − p²r² + p²q + pq² − pqr = p²q² − q²r² − p²r² + 2pqr² + p²q + pq² − pqr
Quick Reference — All Answers at a Glance
| Question | Expression | Answer |
|---|---|---|
| Q1 (i) | (2a − 9)(3a + 4) | 6a² − 19a − 36 |
| Q1 (ii) | (x − 2y)(2x − y) | 2x² − 5xy + 2y² |
| Q1 (iii) | (kl + lm)(k − l) | k²l − kl² + klm − l²m |
| Q1 (iv) | (m² − n²)(m + n) | m³ + m²n − mn² − n³ |
| Q2 (i) | (x + y)(2x − 5y + 3xy) | 2x² − 3xy + 3x²y + 3xy² − 5y² |
| Q2 (ii) | (a − 2b + 3c)(ab² − a²b) | 3a²b² − a³b − 2ab³ + 3ab²c − 3a²bc |
| Q2 (iii) | (mn − kl + km)(kl − lm) | klmn − lm²n − k²l² + kl²m + k²lm − klm² |
| Q2 (iv) | (p³ + q³)(p − 5q + 6r) | p⁴ − 5p³q + 6p³r + pq³ − 5q⁴ + 6q³r |
| Q3 (i) | (x−2y)(y−3x) + (x+y)(x−3y) − (y−3x)(4x−5y) | 10x² − 14xy |
| Q3 (ii) | (m + n)(m² − mn + n²) | m³ + n³ |
| Q3 (iii) | (a−2b+5c)(a−b) − (a−b−c)(2a+3c) + (6a+b)(2c−3a−5b) | −19a² − 3b² + 3c² − 34ab + 16ac |
| Q3 (iv) | (pq−qr+pr)(pq+qr−pr) + pq(p+q−r) | p²q² − q²r² − p²r² + 2pqr² + p²q + pq² − pqr |
Common Mistakes to Avoid in Exercise 11.3
- Partial distribution: In (a + b)(c + d + e), students sometimes only distribute a across all three terms but forget to fully distribute b. Every term in the first bracket must multiply every term in the second bracket without exception.
- Negative × negative errors: In Q1(ii) and Q2(ii), products like (−2y)(−y) and (−2b)(−a²b) both yield positive results. The most common mistake is writing a negative sign for these. Always determine the sign of the product before writing the coefficient.
- Wrong exponent addition with multiple variables: In multi-variable terms, remember to add exponents per variable separately. In Q1(iv), m² × m = m³ (only the m exponents add). Never add exponents across different variables.
- Incorrect sign-flipping when subtracting: In Q3(i) and Q3(iii), when a product is subtracted (−P2), every term inside P2 changes sign when the bracket opens. A very common error is flipping only the first term's sign and leaving the rest unchanged.
- Combining unlike terms: After expanding, terms like m²n and mn² look almost identical but are not like terms — the exponents are on different variables. In Q1(iv), m²n stays separate from mn². Only terms with exactly the same variable-exponent combination can be combined.
What Exercise 11.3 Prepares You For
The binomial multiplication skills developed in Exercise 11.3 are the direct foundation for algebraic identities — the most important shortcuts in school algebra. The result you saw in Q3(ii), where (m + n)(m² − mn + n²) simplified to m³ + n³, is exactly the sum of cubes identity that appears in the Class 9 Polynomials chapter.
The product (a + b)(a − b) = a² − b², (a + b)² = a² + 2ab + b², and (a − b)² = a² − 2ab + b² are all just special cases of the binomial multiplication you have practised here. You will study these identities formally in Exercise 11.3 extensions and later in Class 9 Polynomials.
For students in CBSE, Telangana, and Andhra Pradesh, the "Find the product" and "Simplify" question types from Exercise 11.3 directly parallel the question formats seen in 2-mark and 3-mark problems in Class 8 and Class 9 board examinations. Mastery here pays dividends across all subsequent algebra chapters.