Chapter 11.4 — Exercise 11.3 — Binomial x Binomial
Multiplying a binomial by a binomial or trinomial. This is Lesson 4 of 6 in Chapter 11: Algebraic Expressions.
Every Term Meets Every Term
Multiplying two binomials means every term in the first bracket has to multiply every term in the second — two terms times two terms gives four products before anything gets combined. Finding the product of (2a + 3b) and (a − 3b):
(2a+3b)(a−3b) = 2a×a + 2a×(−3b) + 3b×a + 3b×(−3b)
= 2a² − 6ab + 3ab − 9b² = 2a² − 3ab − 9b²Scaling the Same Rule Up to a Trinomial
A trinomial on one side doesn't change the method, only the term count: (3x − 5y) against (2x² − 3xy + 3y²) needs six products instead of four, since 2 terms × 3 terms = 6:
(3x−5y)(2x²−3xy+3y²) = 6x³−9x²y+9xy²−10x²y+15xy²−15y³
= 6x³ − 19x²y + 24xy² − 15y³Both examples follow the same "every-term-meets-every-term" rule — only the number of resulting products (four versus six) depends on how many terms sit in each bracket. In general, a bracket with m terms multiplied by a bracket with n terms always produces m×n individual products before any combining happens, whether or not any of those products later turn out to be like terms with one another.
Four Binomial Pairs, Worked in Full
Running the same process across four different pairs makes the pattern reliable rather than a one-off trick:
| Pair | Product |
|---|---|
| (2a − 9)(3a + 4) | 6a² − 19a − 36 |
| (x − 2y)(2x − y) | 2x² − 5xy + 2y² |
| (kl + lm)(k − l) | k²l − kl² + klm − l²m |
| (m² − n²)(m + n) | m³ + m²n − mn² − n³ |
The last row is a preview of something more useful than it first looks: (m² − n²)(m + n) doesn't simplify down to fewer terms, because m³, m²n, −mn², and −n³ share no common variable-and-exponent combination with each other — a four-term answer here isn't a sign that something went wrong, it's simply the correct, fully expanded result.
When One Side Has Three Terms Instead of Two
The same distribute-every-term method extends cleanly to a binomial against a trinomial, or a trinomial against a binomial — only the number of products (2×3 = 6 either way) changes:
(x+y)(2x−5y+3xy) = 2x² − 3xy + 3x²y + 3xy² − 5y²
(a−2b+3c)(ab²−a²b) = 3a²b² − a³b − 2ab³ + 3ab²c − 3a²bc
(mn−kl+km)(kl−lm) = klmn − lm²n − k²l² + kl²m + k²lm − klm²
(p³+q³)(p−5q+6r) = p⁴ − 5p³q + 6p³r + pq³ − 5q⁴ + 6q³rThe third row is the easiest to lose track of, since it has the most distinct variables (k, l, m, n) spread across only two factors — writing out all six products before combining any of them, exactly as shown in the two worked examples at the top of this lesson, is what keeps a problem like this from going wrong partway through. The fourth row is worth a second look for a different reason: p³ + q³ and p − 5q + 6r share no variable in common with an obvious pairing, yet the distributive rule still applies exactly as before — every term in the first bracket multiplies every term in the second, regardless of whether the two brackets "look like" they belong together.
Reconstructing a Three-Bracket Sum
A more demanding simplification combines three separate binomial products into a single expression: (x − 2y)(y − 3x) + (x + y)(x − 3y) − (y − 3x)(4x − 5y). Expanding each of the three products before combining anything:
(x−2y)(y−3x) = xy − 3x² − 2y² + 6xy
(x+y)(x−3y) = x² − 3xy + xy − 3y²
−(y−3x)(4x−5y) = −(4xy−5y²−12x²+15xy) = 12x² − 19xy + 5y²Adding all three lines together and collecting like terms:
Total = 10x² − 14xyEvery x² term across the three expansions (−3x², x², 12x²) adds to 10x², every y² term (−2y², −3y², 5y²) cancels down to exactly 0, and the xy terms across all three lines settle at −14xy — a good example of why fully expanding every bracket first, then collecting like terms in one final pass, is more reliable than trying to simplify while brackets are still open.
A Three-Term Product That Collapses to Two Terms
Not every simplification ends in a long expression — some collapse dramatically. Simplifying (m + n)(m² − mn + n²):
(m+n)(m²−mn+n²) = m³ − m²n + mn² + m²n − mn² + n³ = m³ + n³Six products come out of the expansion, but four of them — m²n and −m²n, mn² and −mn² — cancel each other out in pairs, leaving just m³ + n³. This particular binomial-times-trinomial pattern is worth remembering by sight, since it's the standard way of expanding a sum of two cubes without needing to multiply everything out from scratch each time it appears.
Checking the Cancellation With Real Numbers
A quick way to build confidence in a cancellation-heavy result like m³ + n³ is to substitute small numbers on both sides and check they agree. Taking m = 2, n = 1: the original product (m+n)(m²−mn+n²) becomes (2+1)(4−2+1) = 3 × 3 = 9, and the simplified answer m³+n³ becomes 8 + 1 = 9 — the same value, confirming the simplification holds for at least this one case. This substitution check doesn't replace the algebra (it can't prove a simplification is correct for every possible value, only that it hasn't been contradicted by the one value tried), but it's a fast way to catch an obvious sign or arithmetic slip before moving on, especially in a lesson where several terms are expected to cancel out entirely.
A Longer Simplification, Same Careful Process
The most demanding simplification in this exercise combines three separate two-bracket products: (a − 2b + 5c)(a − b) − (a − b − c)(2a + 3c) + (6a + b)(2c − 3a − 5b). Expanding each product in turn and only then collecting like terms across all three results:
Total = −19a² − 3b² + 3c² − 34ab + 16acNotice there's no bc term at all in the final answer — not because bc terms were never produced during the three expansions, but because whatever bc terms did appear across the three products happened to cancel each other out exactly, the same way the y² terms vanished in the three-bracket sum earlier in this lesson. A missing term in a final answer is frequently a sign of genuine cancellation, not a sign that a term was overlooked. It's worth double-checking this the same way as before, by substituting a small set of numbers into both the original three-product expression and the final four-term answer and confirming they agree — a slower check than counting terms by eye, but a far more reliable one when four separate variables are involved across three different brackets at once.
Where Repeated Multiplication Leads
Every product in this lesson was worked out the long way, term by term. Exercise 11.4 shows that a handful of binomial-times-binomial patterns — squaring a sum, squaring a difference, and multiplying a sum by a difference — turn up so often that they're worth memorising as ready-made identities, so the full six-term (or four-term) expansion never has to be repeated from scratch again — the same shortcut that already made m³ + n³ recognisable on sight earlier in this lesson, generalised into a small, reusable toolkit.