Chapter 11.3 — Exercise 11.2 — Monomial x Polynomial

Reaching One Term Further, Into a Trinomial

The same idea extends to a trinomial bracket without any new rule — the monomial just reaches one term further. Finding the product of −5xy and (2x² − 3xy + 3y²):

Both examples follow the identical pattern: however many terms sit inside the bracket, the outside monomial multiplies every one of them exactly once, and every result keeps whatever sign it picks up from multiplying a negative by a positive or a negative by a negative.

Five Products, Filled Into One Table

A table of monomial-times-polynomial products makes the pattern easy to check at a glance, since every row follows the same distribute-and-collect process:

The last row is the one worth slowing down on: with three variables already in the monomial and four spread across the trinomial, it's easy to lose track of an exponent. Taking just the first term as a check — a²b²c² × a²bc — the a's combine as a²×a² = a⁴, the b's as b²×b = b³, and the c's as c²×c = c³, giving a⁴b³c³ exactly as the table shows. The second term brings in a fourth variable, d, that never appeared in the monomial at all: a²b²c² × b²cd simply carries that d straight through into the answer unchanged, since there's nothing in the monomial to combine it with — a reminder that a variable present in only one of the two factors doesn't need "matching," it just tags along into the product exactly as it was.

Simplifying Before Evaluating

4y(3y + 4) simplifies the same way as any monomial-times-binomial product:

A slightly longer version asks for both the simplified form and its value at specific numbers. Simplifying x(2x² − 7x + 3):

At x = 1: 2(1) − 7(1) + 3(1) = 2 − 7 + 3 = −2. At x = 0: every term contains a factor of x, so every term vanishes, leaving 0 — a quick reminder that any polynomial with no constant term always evaluates to zero when x = 0, without needing to substitute and calculate at all. This shortcut only works because of the specific form 2x³ − 7x² + 3x — every single term still carries at least one factor of x after simplifying, which is guaranteed here since the original expression was itself x multiplied by something, and multiplying anything by x = 0 always gives 0 regardless of what that "something" was.

Adding Three Separate Products

Some problems ask for the sum of several monomial-times-binomial products rather than just one. Adding a(a − b), b(b − c), and c(c − a):

A related problem adds x(x + y − r), y(x − y + r), and z(x − y − z):

Both problems follow the same two-step routine — expand every bracket completely first, then hunt for like terms across all three expanded results at once, rather than trying to combine anything while the brackets are still unopened. It's also worth noticing the pattern behind how a(a−b), b(b−c), and c(c−a) were chosen: each variable multiplies a bracket built from itself and the next variable in a repeating cycle (a→b, b→c, c→a). That cyclic structure is exactly why the final answer comes out so symmetric — a² + b² + c² − ab − bc − ac treats all three variables identically, with no single one singled out, which is a useful pattern to recognise since it reappears often whenever a problem is built by cycling the same rule through several variables in turn.

Subtracting One Product From Another

Subtracting the product 2x(5x − y) from the product of 3x(x + 2y):

A second, slightly larger version subtracts 3k(5k − l + 3m) from 6k(2k + 3l − 2m):

Subtraction here is where sign errors creep in most easily, since every term of the second product needs its sign flipped before it can be combined with the first — 18kl minus (−3kl) becomes 18kl + 3kl = 21kl, not 18kl − 3kl, precisely because the bracket being subtracted already contained a negative term. A reliable habit for avoiding this slip is to fully expand both products first, writing each one down on its own line exactly as shown above, and only then subtract term by term — trying to distribute the minus sign and multiply in the same mental step is exactly where a term's sign gets forgotten.

Three Terms Multiplying Three Different Trinomials

The most demanding problem in this exercise multiplies three separate squared monomials against three different trinomials, then combines everything into one simplified expression:

Nine terms come out of the three expansions before any combining happens, and none of them turn out to be like terms with each other — the "simplifying" here is really just organising the nine terms into a readable order (the pure cubes first, then the paired terms), not merging anything down to fewer terms than it started with. That's a useful thing to expect in advance: not every "simplify" instruction results in a shorter expression. Sometimes, as here, the honest simplification is a clearly organised nine-term answer rather than a shorter one that doesn't actually exist, because none of the nine terms genuinely share the same combination of variables and exponents with any other.

Moving From One Bracket to Two

Every product in this lesson kept one side of the multiplication down to a single monomial. Exercise 11.3 removes that restriction, multiplying a binomial by another binomial or trinomial — which means the distributive step from this lesson now has to run twice, once for each term on the left-hand side — the same careful term-by-term bookkeeping practised throughout this exercise, just applied on both sides of the multiplication at once instead of only one.