Multiplication of algebraic expressions — monomial × polynomial, polynomial × polynomial: Worked examples

This page is the practice ground for the parent topic. The parent showed you how to multiply. Here we hunt down every kind of problem the topic can throw at you — every sign pattern, the sneaky "zero" cases, the special-product shortcuts, a real-world word problem, and one exam-style trap — and work each one to the ground.

Before line one: a reminder of the only tool we use.

The picture below is that rule. Two brackets sit on the left and right; every red arrow is one "handshake" — one term of the left bracket meeting one term of the right. Four terms, four arrows, four products. Nothing may be skipped.

Figure — Multiplication of algebraic expressions — monomial × polynomial, polynomial × polynomial

(Alt: two stacked term-lists $a, b$ on the left and $c, d$ on the right; four red arrows connect every left term to every right term, each labelled with its product $a c, a d, b c, b d$ .)

The scenario matrix

Every problem in this topic lands in exactly one of the cells below. First, the quick visual decision map — ask the questions top-down and it drops you into the right case:

(Alt: a decision flowchart. First question "one bracket a single term?" branches to monomial cases A/B; otherwise "same brackets, opposite signs?" branches to difference-of-squares E; "a bracket squared?" to perfect-square D; else the general FOIL/handshake cases C/F, with the degenerate $0$ / $1$ input G flagged as a pre-check.)

The full table (each worked example is labelled with the cell it fills):

#	Case class	What makes it tricky	Example that covers it
A	Monomial × polynomial, all positive	warm-up, no sign traps	Ex 1
B	Monomial × polynomial, negative monomial	sign flips on every term	Ex 2
C	Binomial × binomial, mixed signs	the FOIL sign-tracking trap	Ex 3
D	Special product — perfect square	the missing middle term $2 ab$	Ex 4
E	Special product — difference of squares	middle terms cancel to zero	Ex 5
F	Two-variable trinomial × binomial	many like-term collisions	Ex 6
G	Degenerate: multiply by $0$ / by $1$	limiting cases you must recognise	Ex 7
H	Real-world word problem (area)	translate words → brackets	Ex 8
I	Exam twist — solve for a hidden coefficient	run multiplication backwards	Ex 9

Recall Which cells are the "danger zones"?

Cells B, C, E, G ::: signs (B, C), a case where the middle vanishes (E), and the degenerate $0$ / $1$ inputs (G) that students misread.

The two rules from Exponent Rules we will lean on constantly:

Case A — monomial × polynomial (all positive)

Case B — negative monomial (signs flip everywhere)

Worked example Ex 2 · the negative-out-front case

Multiply $- 5 m^{3} (2 m^{2} - 4 m - 3)$ .

Forecast: The monomial is negative. Predict the sign of each of the three answer terms before computing. (A negative times $+ 2 m^{2}$ ? A negative times $- 4 m$ ? A negative times $- 3$ ?)

Step 1. Distribute $- 5 m^{3}$ : $(- 5 m^{3}) (2 m^{2}) + (- 5 m^{3}) (- 4 m) + (- 5 m^{3}) (- 3)$ Why keep the minus attached? The $- 5 m^{3}$ is a single object; its sign travels with it into every product.

Step 2. Handle sign, then number, then exponent for each:

$(- 5) (+ 2) = - 10$ , $m^{3 + 2} = m^{5}$ → $- 10 m^{5}$
$(- 5) (- 4) = + 20$ , $m^{3 + 1} = m^{4}$ → $+ 20 m^{4}$ (negative × negative = positive)
$(- 5) (- 3) = + 15$ , $m^{3}$ → $+ 15 m^{3}$

Why this step? Missing one sign flip is the classic error — do signs first, deliberately.

✓ Answer: $- 10 m^{5} + 20 m^{4} + 15 m^{3}$

Verify: Set $m = 1$ . Original: $- 5 (2 - 4 - 3) = - 5 (- 5) = 25$ . Answer: $- 10 + 20 + 15 = 25$ . ✓

Case C — binomial × binomial, mixed signs (the FOIL trap)

Worked example Ex 3 · sign-tracking with FOIL

Multiply $(3 x - 4) (2 x - 5)$ .

Forecast: Both brackets have a minus. Will the middle term be positive or negative? Will the constant be positive or negative? Guess now.

Step 1. FOIL — First, Outer, Inner, Last. This is the four-arrow handshake picture again, now with real numbers:

First: $3 x \cdot 2 x = 6 x^{2}$
Outer: $3 x \cdot (- 5) = - 15 x$
Inner: $(- 4) \cdot 2 x = - 8 x$
Last: $(- 4) \cdot (- 5) = + 20$

Why FOIL? It is just "every term × every term" for the $2 \times 2$ case — four handshakes, none skipped.

Step 2. Combine the two middle (like) terms: $6 x^{2} + (- 15 x - 8 x) + 20 = 6 x^{2} - 23 x + 20$ Why combine? $- 15 x$ and $- 8 x$ have identical variable part $x$ , so they add.

✓ Answer: $6 x^{2} - 23 x + 20$

Verify: Set $x = 2$ . Original: $(6 - 4) (4 - 5) = (2) (- 1) = - 2$ . Answer: $6 (4) - 23 (2) + 20 = 24 - 46 + 20 = - 2$ . ✓

Case D — perfect square (don't lose the middle term!)

Case E — difference of squares (middle terms vanish!)

Case F — two-variable trinomial × binomial (many like-term collisions)

Worked example Ex 6 · the big one

Multiply $(3 p - 2 q + 1) (2 p + q)$ .

Forecast: Three terms × two terms = six handshake-products. Some of them will collide and combine. Guess: how many terms survive at the end?

Step 1. Distribute each term of the trinomial across $(2 p + q)$ :

$3 p (2 p + q) = 6 p^{2} + 3 pq$
$- 2 q (2 p + q) = - 4 pq - 2 q^{2}$
$1 (2 p + q) = 2 p + q$

Why split this way? Treating each of $3 p, - 2 q, 1$ as a mini-monomial keeps the bookkeeping clean — this is just distribution used three times.

Step 2. Write all six products together: $6 p^{2} + 3 pq - 4 pq - 2 q^{2} + 2 p + q$

Step 3. Combine like terms. The only collision is the $pq$ pair: $3 pq - 4 pq = - pq$ Why only that pair? $p^{2}$ , $q^{2}$ , $p$ , $q$ each appear once — nothing to combine with.

✓ Answer: $6 p^{2} - pq - 2 q^{2} + 2 p + q$ (five terms survive)

Verify: Set $p = 1, q = 1$ . Original: $(3 - 2 + 1) (2 + 1) = 2 \cdot 3 = 6$ . Answer: $6 - 1 - 2 + 2 + 1 = 6$ . ✓

Case G — degenerate inputs (multiply by $0$ and by $1$ )

You must recognise these instantly, because they look like they need work but don't.

Case H — real-world word problem (area model)

Worked example Ex 8 · the widening garden

A rectangular garden is $(x + 3)$ metres long and $(x + 2)$ metres wide. A path widens both dimensions by $2$ m. Write the new area as an expanded polynomial.

Forecast: New length and width each grow by $2$ . What are the new side lengths?

Step 1. New length $= (x + 3) + 2 = x + 5$ ; new width $= (x + 2) + 2 = x + 4$ . Why? "Widens by $2$ " adds $2$ to each side length.

Step 2. Area of a rectangle is length × width. The area model below cuts the big rectangle into four pieces — one for each handshake-product:

(Alt: a rectangle of width $x + 5$ and height $x + 4$ , split by one vertical and one horizontal line into four sub-rectangles. The top-left square is filled red and labelled $x^{2}$ ; the top-right strip is $5 x$ ; the bottom-left strip is $4 x$ ; the small bottom-right corner square is $20$ .)

Read the picture piece by piece:

the red top-left square has sides $x$ by $x$ , so its area is $x \cdot x = x^{2}$ ;
the top-right strip has sides $5$ by $x$ , area $5 x$ ;
the bottom-left strip has sides $x$ by $4$ , area $4 x$ ;
the small bottom-right corner square has sides $5$ by $4$ , area $20$ .

$Area = (x + 5) (x + 4)$

Step 3. Add the four sub-areas (the two strips are the like terms): $x^{2} + 5 x + 4 x + 20$

Step 4. Combine: $5 x + 4 x = 9 x$ .

✓ Answer: $Area = x^{2} + 9 x + 20$ square metres.

Verify (with units): Take $x = 1$ m. New sides: $6$ m by $5$ m, area $= 30 m^{2}$ . Formula: $1 + 9 + 20 = 30 m^{2}$ . ✓ Units are m × m = m², correct for area.

Case I — exam twist (run the multiplication backwards)

Recall Self-test: name the cell, then solve

$(2 x - 9) (2 x + 9)$ — which cell? ::: Cell E (difference of squares) → $4 x^{2} - 81$ . $- 3 t^{2} (t - 5)$ — which cell? ::: Cell B (negative monomial) → $- 3 t^{3} + 15 t^{2}$ . $(x + 6)^{2}$ — which cell? ::: Cell D (perfect square) → $x^{2} + 12 x + 36$ .

Related build-outs: Quadratic Expressions (every binomial × binomial lands here), Pascal's Triangle (for $(a + b)^{n}$ coefficients), and running this all backwards → Factoring Polynomials.