Exercise 11.2 — Monomial x Polynomial
Multiplying a binomial or trinomial by a monomial.
What is Exercise 11.2 About?
Exercise 11.2 of Class 8 Mathematics (CBSE, Telangana & Andhra Pradesh syllabus) introduces the multiplication of a monomial by a polynomial — specifically by a binomial (two terms) or a trinomial (three terms). This is a natural extension of Exercise 11.1, where you only multiplied single terms together.
The key technique used throughout this exercise is the distributive law: when a monomial is multiplied by a bracket containing two or more terms, the monomial is distributed and multiplied with each term inside the bracket individually. The resulting terms are then added (keeping signs carefully).
a × (b + c) = a×b + a×c ← Distributive Law
Understanding the Method — Two Introductory Examples
Before the exercise questions begin, the textbook demonstrates the distributive technique with two examples. Understanding these examples makes every question in Exercise 11.2 straightforward.
Find the product of 4a²b and (a − 3b)
The monomial 4a²b is multiplied separately with a (giving 4a³b) and with −3b (giving −12a²b²). The two results are then written as a sum.
Find the product of −5xy and (2x² − 3xy + 3y²)
Here −5xy is distributed across all three terms. Notice that (−5xy) × (−3xy) = +15x²y² — a negative times a negative gives a positive.
Question 1 — Complete the Product Table
This question provides a table where one expression is a polynomial (binomial or trinomial) and the other is a monomial. You must multiply them using the distributive law and fill in the product column.
| S.No. | First Expression | Second Expression | Product |
|---|---|---|---|
| (i) | 5q | p + q − 2r | 5pq + 5q² − 10qr |
| (ii) | kl + lm + mn | 3k | 3k²l + 3klm + 3kmn |
| (iii) | ab² | a + b² + c³ | a²b² + ab⁴ + ab²c³ |
| (iv) | x − 2y + 3z | xyz | x²yz − 2xy²z + 3xyz² |
| (v) | a²bc + b²cd − abd² | a²b²c² | a⁴b³c³ + a²b⁴c³d − a³b³c²d² |
Let's trace row (ii) step by step to see the method clearly:
And row (v) — the most complex row — step by step:
Question 2 — Simplify 4y(3y + 4)
This is a straightforward application of the distributive law on a monomial times a binomial.
4y × 3y: coefficients → 4 × 3 = 12; variables → y × y = y². Result: 12y².
4y × 4: coefficients → 4 × 4 = 16; variable → y. Result: 16y.
Question 3 — Simplify and Find Values for x = 1 and x = 0
This two-part question first asks you to expand the expression by distributing the monomial, then substitute specific values of x to evaluate the result. It tests both algebraic manipulation and numerical substitution.
Step 1 — Expand the expression
Step 2 — Evaluate by substitution
= 2(1) − 7(1) + 3
= 2 − 7 + 3
= 2(0) − 7(0) + 0
= 0 − 0
Questions 4 & 5 — Add the Products
These questions require you to expand each expression separately using the distributive law, then add all the results together and simplify by collecting like terms.
Expand each product
Add all three results
No further simplification is possible — there are no like terms to collect.
Expand each product
Add all three and collect like terms
Questions 6 & 7 — Subtract the Products
Subtraction problems add one extra layer: after expanding both expressions, the second expanded result is placed inside a bracket with a minus sign in front. When you open that bracket, every sign inside flips. This is where most mistakes happen.
Expand both products
Subtract — flip all signs of the second result
Expand both products
Subtract — flip all signs of the second result
Question 8 — Simplify the Large Expression
This is the most complex question in Exercise 11.2. It requires distributing three separate monomials (a², b², and −c²) across their respective brackets, then adding all three expanded results together and collecting like terms.
Expand each bracket separately
Add all three results
Quick Reference — All Answers at a Glance
| Question | Expression | Answer |
|---|---|---|
| Q1(i) | 5q × (p + q − 2r) | 5pq + 5q² − 10qr |
| Q1(ii) | 3k × (kl + lm + mn) | 3k²l + 3klm + 3kmn |
| Q1(iii) | ab² × (a + b² + c³) | a²b² + ab⁴ + ab²c³ |
| Q1(iv) | xyz × (x − 2y + 3z) | x²yz − 2xy²z + 3xyz² |
| Q1(v) | a²b²c² × (a²bc + b²cd − abd²) | a⁴b³c³ + a²b⁴c³d − a³b³c²d² |
| Q2 | 4y(3y + 4) | 12y² + 16y |
| Q3 (expand) | x(2x² − 7x + 3) | 2x³ − 7x² + 3x |
| Q3 (x=1) | 2(1)³ − 7(1)² + 3(1) | −2 |
| Q3 (x=0) | 2(0)³ − 7(0)² + 3(0) | 0 |
| Q4 | a(a−b) + b(b−c) + c(c−a) | a² + b² + c² − ab − bc − ac |
| Q5 | x(x+y−r) + y(x−y+r) + z(x−y−z) | x² − y² − z² + 2xy − yz + xz − xr + yr |
| Q6 | 3x(x+2y) − 2x(5x−y) | −7x² + 8xy |
| Q7 | 6k(2k+3l−2m) − 3k(5k−l+3m) | −3k² + 21kl − 21km |
| Q8 | a²(a−b+c) + b²(a+b−c) − c²(a−b−c) | a³+b³+c³ − a²b+ab² − b²c+bc² + a²c−ac² |
Common Mistakes to Avoid
- Distributing to only the first term: In a(b + c + d), students sometimes write only ab + c + d — forgetting to multiply the monomial with every term. Always check: number of terms in your answer = number of terms in the bracket.
- Sign errors when subtracting: In Q6 and Q7, opening the minus bracket requires flipping all signs inside. The most common error is flipping only the first sign. Write the bracket step explicitly before simplifying.
- Forgetting negative × negative = positive: In Q3 — x × (−7x) = −7x², not +7x². Always identify and handle the sign before multiplying.
- Incorrect exponent addition in multi-variable terms: In Q1(v), multiplying a²b²c² by a²bc requires adding exponents for each variable separately: a²⁺² = a⁴, b²⁺¹ = b³, c²⁺¹ = c³. Mixing up which exponents go with which variable is a frequent slip.
- Combining unlike terms: After expanding, terms like a²b and ab² look similar but are not like terms — the exponents are on different variables. Only combine terms that are identical in both variable and exponent.
(3x² + 6xy) − (10x² − 2xy) as a separate step. Examiners give marks for this intermediate step.
What Exercise 11.2 Prepares You For
The distributive method you practise in Exercise 11.2 is the direct foundation for Exercise 11.3, which covers multiplying two binomials together — a skill that leads directly to the famous algebraic identities in Class 9.
Questions 4 and 5 (adding products) and Questions 6 and 7 (subtracting products) are early forms of the expansion and simplification problems you will encounter extensively in Class 9 Polynomials. The sign-handling discipline developed here is essential throughout secondary school mathematics.
For students in CBSE, Telangana, and Andhra Pradesh, the "simplify and evaluate" style of Question 3 — expand first, then substitute — appears in 2-mark and 3-mark questions in board exams, making it one of the most important question types in Chapter 11.