Chapter Summary and Practice

A few weeks after learning a method, it is easy to remember the formula but forget why it works. This lesson brings the chapter back together. You will move from multiplying parts to recognising special products, checking arguments and interpreting patterns. The aim is to make a sensible choice when a question does not tell you which identity to use. You should also be able to explain each important step to another student.

Begin by trying to recall the main ideas without looking at the formulas. What must an outside factor multiply? Why does a squared sum have a middle term? How can two different expressions count the same pattern? Use the recap to fill any gaps, then attempt the worked examples and mixed questions.

The closing coin puzzle asks a related question in a different setting: how can you rearrange something while preserving as much of it as possible? Finding a short solution is one achievement. Explaining why no shorter solution can work requires an additional argument.

Summary

Distribution means multiplying every signed term. For example, 7(20 + 3) = 7 × 20 + 7 × 3. The same reasoning applies to subtraction and to expressions with more than two terms. With two brackets, every term in one bracket must meet every term in the other.

An identity is an equality true for every allowed value of its letters. A valid expansion establishes an identity because each step preserves the value generally. Substitution is useful for catching mistakes, but several matching examples are still only checks. A single mismatch is enough to show that a claimed identity is false.

When ab becomes (a + m)(b + n), the new product is ab + an + bm + mn. Subtract ab to find the change an + bm + mn. Negative increments describe decreases. Do not assume the change is positive just because one or both factors have increased.

When both factors increase by 1, the change is a + b + 1. When the first increases by 1 and the second decreases by 1, it is b − a − 1. The extra product of the increments accounts for the corner in an area model. Forgetting it is the same kind of error as omitting a region from a diagram.

The three main identities are special cases of distribution. A square of a sum has two rectangles of area ab alongside its two squares. A square of a difference can be calculated by removing two strips and restoring their overlap once. A difference of squares can be rearranged into a rectangle whose sides are the sum and difference of the original sides.

The pictures use positive lengths. The algebraic proofs apply more widely, including negative integers and fractions. Treat a whole term such as 6x or 3z/4 as one part when substituting: (6x)² = 36x² and (3z/4)² = 9z²/16. Squaring changes both the coefficient and the variable power.

Like terms have the same variables with the same powers. Thus 2a²b + 5ba² = 7a²b, but a²b + ab² remains a sum of unlike terms. Reordering factors does not change a term; changing powers does. A long expansion should first show every product and only then collect matching terms.

This also explains larger expressions. Distributing (a + b)(a² + 2ab + b²) and collecting like terms gives a³ + 3a²b + 3ab² + b³. In products such as (a − b)(a² + ab + b²), the interior terms cancel, leaving a³ − b³. Both results use the same basic property, even though they contain higher powers.

Fast multiplication uses convenient splits. Multiply by 11 using 10N + N, by 101 using 100N + N, and by 1001 using 1000N + N. Keep place values aligned and carry as needed. Multiplication by 99 or 999 uses 100N − N or 1000N − N.

For mental squares, choose a nearby convenient number and use a squared sum or difference. For products equally spaced from a common centre, use a difference of squares. You can also rearrange that identity as a² = (a + b)(a − b) + b² to turn a square into an easy product plus a small correction.

In a pattern problem, identify the stage variable and describe the pieces before writing a formula. The missing-corner circle pattern has k(k + 2) = k² + 2k = (k + 1)² − 1 circles. A one-tile border around an n by n hole has (n + 2)² − n² = 4n + 4 tiles.

The yellow stepped pattern has y(y + 2) + 2(y + 2) = (y + 2)² tiles. The blue pattern has (y + 1)² + y tiles. A sequence of totals alone is not a complete explanation: the arrangement shows why each term belongs.

Area methods may subtract known pieces from a whole or directly multiply the sides of what remains. Four m by n rectangles inside a square give the central area (m + n)² − 4mn = (n − m)². Three x by y rectangles in the other arrangement leave x² − xy = x(x − y). For two crossing removed strips, add the overlap once to correct double subtraction.

Algebra can prove relationships that a few numerical examples only suggest. In a calendar block, (a + 1)(a + 7) − a(a + 8) = 7. For three consecutive integers n − 1, n, n + 1, the middle square exceeds the outer product by 1.

An even integer 2t has square 4t². An odd integer 2t + 1 has square 4t(t + 1) + 1; because one of t and t + 1 is even, this is 1 more than a multiple of 8. If A and B have remainders 3 and 5 on division by 7, their sum and product each have remainder 1. A − B has remainder 5, whereas B − A has remainder 2.

Language and dimensions deserve the same care as signs. “Two more than a square” is s² + 2, and “half the square of a sum” is (a + b)²/2. In the two-lawn park layout, outer margins w and a central gap 2w give tiled area 8w(g + w). A different central gap would change the expression.

It’s Puzzle Time!

Place equal coins at equally spaced points in an upright triangle. A triangle with two rows contains 1 + 2 = 3 coins, one with three rows contains 6, and one with four rows contains 10. The next triangle has five rows and 15 coins.

You may pick up one coin and place it at a new position; this counts as one move. Keep all coins, keep the same spacing in the completed triangle, and make it point downward. The final triangle may be shifted on the table. You are not required to keep its outline in the original location.

Coin Conjoin

Before moving anything, imagine the final downward triangle lying over the original one. Every position shared by both triangles can keep its coin. Only coins outside the shared positions need to move to new positions. Therefore a good strategy is to seek the greatest overlap.

For 3 coins, move the top coin to a position below the centre of the bottom pair. The pair becomes the top row of a downward triangle. Exactly one coin moves. Zero moves cannot change the orientation, so one is minimal.

For 6 coins, keep the top coin, the two coins in the middle row, and the centre coin of the bottom row. Move the two end coins of the bottom row to the left and right of the top coin, at the same spacing. The final rows contain 3, 2 and 1 coins. At most four positions can overlap for this size, so two moves are necessary as well as sufficient.

Why can two moves not be enough? It would require eight coins to stay in place. Compare the original row lengths 1, 2, 3, 4 with the downward row lengths 4, 3, 2, 1. For any vertical shift, the shared coins in a row cannot exceed the shorter of the two rows.

The best row alignment puts the four-coin target row level with the original two-coin row. The row limits are then 2, 3 and 2, totalling 7. Other vertical alignments give limits of at most 6. A horizontal shift cannot increase any row beyond its shorter row length. Therefore at most seven coins stay and at least three must move. The construction achieves that limit.

For example, shift the target top row two rows down. Its rows of 5, 4 and 3 coins meet original rows of 3, 4 and 5 coins. The limits are 3, 4 and 3, totalling 10. The two remaining target rows have no original coins beneath them. Shifting farther than four rows leaves no shared row at all.

The distinction is important: a construction shows that a particular number of moves is possible; the overlap limit shows that fewer moves are impossible. Together, they establish a minimum. The same distinction appears in algebra when a numerical check suggests a claim and a general argument proves it.