Multiplication of algebraic expressions — monomial × polynomial, polynomial × polynomial: Exercises

This is your training gym for the parent topic. Work each problem on paper before opening the solution. The problems climb a staircase: first you just recognise the pattern, and by the end you invent your own.

Before we start, one picture to remind you what "multiply two sums" looks like — it is an area (a rectangle chopped into smaller rectangles), which is the Area Models for Algebra way of seeing it.

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

Look at the grid: the big rectangle has width $(x + 2)$ and height $(x + 3)$ . Its total area is the product $(x + 2) (x + 3)$ . But the four coloured tiles inside are the four "handshakes" — $x \cdot x$ , $x \cdot 3$ , $2 \cdot x$ , $2 \cdot 3$ . Adding the tiles = doing the multiplication. That's the whole chapter in one image.

Level 1 — Recognition

Here you only need to spot "monomial times a bunch of terms" and hand each term its share.

Recall Solution 1.1

What we do: give the $5$ to each term inside. $5 (x + 4) = 5 \cdot x + 5 \cdot 4 = 5 x + 20$ Why: the parentheses are a "package deal" — $5$ must touch both the $x$ and the $4$ . ✓ Answer: $5 x + 20$

Recall Solution 1.2

Distribute $3 x$ to each term: $3 x \cdot 2 x + 3 x \cdot 7$ Multiply the number parts, add the exponents on like bases (Exponent Rules): $x^{1} \cdot x^{1} = x^{1 + 1} = x^{2}$ . $= 6 x^{2} + 21 x$ ✓ Answer: $6 x^{2} + 21 x$

Recall Solution 1.3

Hand $- 4 a^{2}$ to all three terms, tracking signs carefully.

$(- 4 a^{2}) (a^{2}) = - 4 a^{2 + 2} = - 4 a^{4}$
$(- 4 a^{2}) (- 2 a) = + 8 a^{2 + 1} = 8 a^{3}$ (negative × negative = positive)
$(- 4 a^{2}) (6) = - 24 a^{2}$ $= - 4 a^{4} + 8 a^{3} - 24 a^{2}$ ✓ Answer: $- 4 a^{4} + 8 a^{3} - 24 a^{2}$

Level 2 — Application

Now two full groups meet. Use FOIL for binomial × binomial, or the "every-with-every" rule for anything bigger.

Recall Solution 2.1

First $x \cdot x = x^{2}$ ; Outer $x \cdot 2 = 2 x$ ; Inner $5 \cdot x = 5 x$ ; Last $5 \cdot 2 = 10$ . $x^{2} + 2 x + 5 x + 10$ Combine the like terms $2 x + 5 x = 7 x$ : $= x^{2} + 7 x + 10$ ✓ Answer: $x^{2} + 7 x + 10$

Recall Solution 2.2

F: $3 x \cdot 2 x = 6 x^{2}$
O: $3 x \cdot 5 = 15 x$
I: $- 4 \cdot 2 x = - 8 x$
L: $- 4 \cdot 5 = - 20$ $6 x^{2} + 15 x - 8 x - 20$ Combine $15 x - 8 x = 7 x$ : $= 6 x^{2} + 7 x - 20$ ✓ Answer: $6 x^{2} + 7 x - 20$

Recall Solution 2.3

Binomial × trinomial: give each term of $(x + 2)$ to the whole trinomial. $x (x^{2} - 3 x + 4) = x^{3} - 3 x^{2} + 4 x$ $2 (x^{2} - 3 x + 4) = 2 x^{2} - 6 x + 8$ Add and combine like terms: $x^{3} + (- 3 x^{2} + 2 x^{2}) + (4 x - 6 x) + 8 = x^{3} - x^{2} - 2 x + 8$ ✓ Answer: $x^{3} - x^{2} - 2 x + 8$

Level 3 — Analysis

Special products appear. Recognise the shape so you can shortcut — but understand why the shortcut is true.

Recall Solution 3.1

A square means the group times itself: $(x + 6) (x + 6)$ . Use $(a + b)^{2} = a^{2} + 2 ab + b^{2}$ with $a = x$ , $b = 6$ : $x^{2} + 2 (x) (6) + 6^{2} = x^{2} + 12 x + 36$ The $2 ab$ appears because the two "cross" handshakes ( $x \cdot 6$ and $6 \cdot x$ ) are identical and add up. ✓ Answer: $x^{2} + 12 x + 36$

Recall Solution 3.2

Use $(a - b)^{2} = a^{2} - 2 ab + b^{2}$ with $a = 2 y$ , $b = 5$ : $(2 y)^{2} - 2 (2 y) (5) + 5^{2} = 4 y^{2} - 20 y + 25$ ✓ Answer: $4 y^{2} - 20 y + 25$

Recall Solution 3.3

This is the difference of squares shape $(a + b) (a - b) = a^{2} - b^{2}$ , with $a = 3 x$ , $b = 7$ . The Outer and Inner handshakes ( $3 x \cdot (- 7)$ and $7 \cdot 3 x$ ) cancel, leaving only squares: $(3 x)^{2} - 7^{2} = 9 x^{2} - 49$ ✓ Answer: $9 x^{2} - 49$

Level 4 — Synthesis

Combine several moves in one problem, mixed variables, and cleanup at the end.

Recall Solution 4.1

Distribute each of the three terms of the trinomial over $(a + 4 b)$ .

$2 a (a + 4 b) = 2 a^{2} + 8 ab$
$- 3 b (a + 4 b) = - 3 ab - 12 b^{2}$
$1 (a + 4 b) = a + 4 b$ Combine everything; the only like terms are $8 ab - 3 ab = 5 ab$ : $2 a^{2} + 5 ab - 12 b^{2} + a + 4 b$ ✓ Answer: $2 a^{2} + 5 ab - 12 b^{2} + a + 4 b$

Recall Solution 4.2

First piece is difference of squares: $(x + 3) (x - 3) = x^{2} - 9$ . Second piece is a square of a difference: $(x - 4)^{2} = x^{2} - 8 x + 16$ . Add them: $(x^{2} - 9) + (x^{2} - 8 x + 16) = 2 x^{2} - 8 x + 7$ ✓ Answer: $2 x^{2} - 8 x + 7$

Recall Solution 4.3

Trinomial × trinomial = $3 \times 3 = 9$ handshakes. Distribute each left term.

$x^{2} (x^{2} - x + 3) = x^{4} - x^{3} + 3 x^{2}$
$2 x (x^{2} - x + 3) = 2 x^{3} - 2 x^{2} + 6 x$
$- 1 (x^{2} - x + 3) = - x^{2} + x - 3$ Stack by degree and combine: $x^{4} + (- 1 + 2) x^{3} + (3 - 2 - 1) x^{2} + (6 + 1) x - 3$ $= x^{4} + x^{3} + 0 x^{2} + 7 x - 3 = x^{4} + x^{3} + 7 x - 3$ ✓ Answer: $x^{4} + x^{3} + 7 x - 3$

Level 5 — Mastery

Reverse the machine, or design a problem to a target. This is where multiplication meets Factoring Polynomials.

Recall Solution 5.1

Expanding gives $x^{2} + (b + c) x + b c$ . Match coefficients:

$b + c = 9$
$b c = 20$ Two numbers that add to $9$ and multiply to $20$ : $b = 4$ , $c = 5$ (or swapped). $(x + 4) (x + 5) = x^{2} + 9 x + 20 ✓$ ✓ Answer: $b = 4, c = 5$

Recall Solution 5.2

Area $=$ width $\times$ length. FOIL $(2 x + 1) (3 x + 4)$ :

F: $6 x^{2}$ ; O: $8 x$ ; I: $3 x$ ; L: $4$ $= 6 x^{2} + 11 x + 4$ Now substitute $x = 2$ : $6 (4) + 11 (2) + 4 = 24 + 22 + 4 = 50$ Check directly: width $= 5$ , length $= 10$ , so area $= 50$ . ✓ ✓ Answer: Area $= 6 x^{2} + 11 x + 4$ ; at $x = 2$ it is $50$ square units.

Recall Solution 5.3

Notice $103 = 100 + 3$ and $97 = 100 - 3$ , so with $a = 100$ , $b = 3$ : $103 \times 97 = 10 0^{2} - 3^{2} = 10000 - 9 = 9991$ The same "cross terms cancel" idea that killed the middle term in Problem 3.3 gives a lightning mental-math trick. ✓ Answer: $9991$

Recall Self-check: what was the single idea behind all 13 problems?

Every term on the left shakes hands with every term on the right, then you combine like terms — with special products (perfect square, difference of squares) being fast-recognised shortcuts of that same rule.

The distributive rule, its FOIL corners, and its area picture

are three views of one operation: multiply every-with-every, then tidy.

Why does

(a + b) (a - b)

have no middle term?

the Outer and Inner cross products

+ ab

and

- ab

cancel exactly.

In reverse (factoring), matching

(x + b) (x + c) = x^{2} + S x + P

needs which two conditions?

b + c = S

and

b c = P

— both, not either.