We Distribute, Yet Things Multiply | CBSE Class 8 Maths Notes

This note covers distributivity, changes in products, multiplication of algebraic expressions, like terms, quick multiplication, squares of sums and differences, the difference of squares, algebraic mistakes, number patterns, and different methods of counting and finding areas.

What does the distributive property mean?

Algebra uses letters to represent numbers and express patterns and relations. A variable, or letter-number, stands for a number. Here, a, b and c represent numbers. An expression combines numbers or variables using mathematical operations.

The symbols +, −, × and = mean addition, subtraction, multiplication and equality. Brackets group an expression. Writing ab means a × b; similarly, a(b + c) means a multiplied by the whole bracket. A product is the result of multiplication.

Property: Multiplication distributes over addition

a(b + c) = ab + ac.

A term is a part joined to other parts by addition or subtraction, with its sign included. In ab + ac, the terms are ab and ac. Neither term may be left out when the bracket is expanded.

The number a multiplies both b and c, and the resulting products are added. This is the distributive property. Expanding an expression means carrying out the multiplication across the bracket to write it as a sum of terms.

Two adjacent dot arrays have a rows each. Their column counts are b and c, and their totals are labelled ab and ac. Together they represent a(b + c).

Reference: NCERT Class 8, page 137

Why can the bracket come first?

The commutative property of multiplication means that changing the order of the numbers being multiplied does not change their product. Thus, (a + b)c = c(a + b) = ca + cb = ac + bc.

The same reasoning works with integers, meaning positive and negative whole numbers and zero. Algebra lets us express this relationship generally, instead of checking separate numerical multiplications one at a time.

How does a product change when its factors change?

Factors are the numbers or expressions being multiplied. Take a and b as the original factors, with product ab. The change in the product means the new product minus the original product.

What happens when one factor increases?

If b increases by 1, then a(b + 1) = ab + a. The change is a. If a increases by 1 instead, (a + 1)b = ab + b, so the change is b.

Worked example 1. In 23 × 27, increase the second factor by 1. Find the change without multiplying the original factors fully.

Answer: 23(27 + 1) = 23 × 27 + 23. The product therefore increases by 23, because one extra group of 23 is included.

What happens when both factors change?

When both increase by 1, expand one bracket and then the other: (a + 1)(b + 1) = (a + 1)b + (a + 1)1 = ab + b + a + 1. The change is a + b + 1.

When a increases by 1 and b decreases by 1, the new product is (a + 1)(b − 1) = ab + b − a − 1. Its change is b − a − 1; this need not be positive.

Worked example 2. Starting with 23 × 27, increase the first factor by 1 and decrease the second by 1.

Answer: (23 + 1)(27 − 1) = 23 × 27 + 27 − 23 − 1. The change is 3, so this product increases by 3.

The formulas remain valid when letter-numbers take negative integer values. Therefore, describe the result as a change until its sign is known; changing a factor upwards does not by itself settle the direction of the product's change.

How do we multiply two expressions with two terms each?

An identity states that two algebraic expressions are equal for all allowed values of their letter-numbers. It records a general relationship. The distributive property produces identities because its reasoning does not depend on choosing particular values.

Let m and n represent changes to the original factors a and b. The new factors are a + m and b + n. A negative change represents a decrease, so the same method handles increases and decreases.

Identity: The general product of two sums

(a + m)(b + n) = ab + mb + an + mn.

1. Treat the whole first bracket as one quantity.
2. Multiply it by b and by n: (a + m)b + (a + m)n.
3. Expand these products to obtain ab + mb + an + mn.
4. Compare with ab: the change is mb + an + mn.

Every term of one bracket multiplies every term of the other. The extra product mn is essential: it accounts for the interaction of the two changes. Using only the changes mb and an leaves the expansion incomplete.

How are negative changes included?

Worked example 3. Decrease the first factor a by 2 and increase the second factor b by 3. Express the new product and its change.

Answer: (a − 2)(b + 3) = ab − 2b + 3a − 6. Subtracting the original product ab leaves the change 3a − 2b − 6.

If both factors decrease, by 3 and 4 respectively, then (a − 3)(b − 4) = ab − 3b − 4a + 12. The final term is positive because it comes from multiplying two negative terms.

This method is more general than a rule about increases alone. Keep the sign attached to each term, form every product, and only then combine terms that can be combined.

How are longer expressions expanded and like terms collected?

The distributive property is not restricted to two terms within a bracket. A factor outside a bracket must multiply every term inside it, including negative terms and fractions. A fraction represents a division: 3/2 means 3 divided by 2.

The notation a² means a × a and is called the square of a. Similarly, a³ means a × a × a and is its cube. The superscript is an exponent, indicating the number of repeated factors here.

Which terms can be combined?

Like terms have the same variable part, including the same exponents. Their numerical multipliers, called coefficients, can be added. Thus ab and ba are like terms because multiplication is commutative, giving ab + ba = 2ab.

Worked example 4. Expand (3a/2)(a − b + 1/5), where a and b are numbers.

Answer: Multiply 3a/2 by each term. The three products are (3/2)a², −(3/2)ab and (3/10)a. Hence the expansion is (3/2)a² − (3/2)ab + (3/10)a. Their variable parts differ, so they cannot be combined into one term.

How does the method extend to two brackets?

For (a + b)(a² + 2ab + b²), multiply each term of the second bracket by both a and b. Before collecting terms, the expansion is a³ + a²b + 2a²b + 2ab² + ab² + b³.

Now a²b + 2a²b = 3a²b, and 2ab² + ab² = 3ab². The simplified result is a³ + 3a²b + 3ab² + b³. The terms a²b and ab² remain separate because their exponents belong to different letters.

Expansion and collection are different steps. First produce all the products; then identify matching variable parts. Combining unlike terms as though their variable parts matched does not preserve the expression in general.

How can distributivity make numerical multiplication quicker?

Splitting a factor into a convenient sum or difference can replace a difficult multiplication with simpler ones. Multiplying by 11 uses 10 + 1; multiplying by 101 uses 100 + 1. In each case the two resulting products are added.

What explains the rule for multiplying by 11?

Place value is the value a digit has because of its position, such as units, tens, hundreds or thousands. Multiplication by 10 moves each digit one place towards a higher value; the unshifted number is then added underneath. A carry transfers a complete group of ten from one column to the next.

Worked example 5. Find 3874 × 11 using distributivity.

Answer: 3874 × 11 = 3874(10 + 1) = 38740 + 3874 = 42614. Writing the addition by place value explains why neighbouring digits are added in the one-line method.

Work from the units column towards the left. Write 4; then 4 + 7 gives 11, so write 1 and carry 1. Next, 7 + 8 + 1 gives 16; write 6 and carry 1.

The next column gives 8 + 3 + 1 = 12; write 2 and carry 1. Finally, 3 + 1 = 4.

What changes for 101, 1001 and 99?

Multiplication by 101 adds the original number to its product with 100. Multiplication by 1001 similarly adds it to its product with 1000. Correct column alignment matters because these shifts are different from the shift used for 11.

Worked example 6. Calculate 9734 × 99 by writing 99 as 100 − 1.

Answer: 9734(100 − 1) = 973400 − 9734 = 963666. Here the correction is a subtraction because 99 is one less than 100.

Why does the square of a sum contain a middle term?

The square of a sum means multiplying the whole sum by itself. It does not mean squaring each term separately and stopping there. Both terms in the first bracket must multiply both terms in the second bracket.

Identity: Square of a sum

(a + b)² = a² + 2ab + b².

Expanding (a + b)(a + b) gives a² + ab + ba + b². These cross products, ab and ba, each multiply different terms from the two brackets. They are equal, so their sum is 2ab. This explains both the coefficient 2 and the presence of both variables in the middle term.

Area measures the surface inside a shape, in square units. A square's area is its side multiplied by itself. A rectangle's area is its length multiplied by its breadth. These rules give a geometric interpretation of the identity.

Each side of the square is split into lengths 60 and 5. The parts are a square labelled 60², a square labelled 5², and two rectangles labelled 60 × 5 and 5 × 60.

Reference: NCERT Class 8, page 145

Worked example 7. Find the area of a square with side 65 units by splitting the side into 60 + 5.

Answer: 65² = 60² + 2 × 60 × 5 + 5² = 3600 + 600 + 25 = 4225 square units. The two rectangular regions contribute the middle term.

How does this work with an algebraic term?

Let x represent a number. In (6x + 5)², the two terms being squared are 6x and 5. Therefore, (6x + 5)² = (6x)² + 2(6x)(5) + 5² = 36x² + 60x + 25.

The entire term 6x is squared, including its coefficient. The area picture uses positive lengths, while the algebraic expansion follows from distributivity and also applies to negative integer values of the letters.

How do we expand the square of a difference?

A difference is the result of subtraction. To square a − b, multiply a − b by itself. The negative sign belongs to b in each bracket, so it must be included when forming the four products.

Identity: Square of a difference

(a − b)² = a² − 2ab + b².

The expansion is a² − ab − ba + b². The middle terms combine to −2ab. The final term has a plus sign because (−b)(−b) = b². Another route is to regard a − b as a + (−b) and use the square-of-a-sum identity.

Why is the small square added back?

A square of side 55 sits inside a square of side 60. The right and bottom strips each have width 5. The corner where the strips overlap is a square of side 5.

Reference: NCERT Class 8, page 146

Removing both rectangles of dimensions 60 and 5 subtracts the corner square twice. Add its area back once to leave it subtracted only once. This gives 60² − 60 × 5 − 5 × 60 + 5².

Worked example 8. Find 55² using 60² = 3600 and 5² = 25.

Answer: 55² = (60 − 5)² = 3600 − 300 − 300 + 25 = 3025. Thus a square of side 55 units has area 3025 square units.

Keep the square of a difference separate from a difference of squares. The former contains a middle term; the latter is a subtraction between two complete squares. Brackets identify which quantity is being squared.

Reversing the subtraction does not change its square: (a − b)² and (b − a)² both expand to a² − 2ab + b². This equality follows by expansion, even though the unsquared differences have opposite signs.

How does the difference of two squares simplify multiplication?

The difference of squares is an expression such as a² − b². It can be written as a product involving the sum and difference of the same two numbers. This is useful when two factors lie equally far on either side of a convenient number.

Identity: Product of a sum and a difference

(a + b)(a − b) = a² − b².

Distributing gives a² − ab + ba − b². The middle terms cancel, meaning their sum is zero, because ba = ab. Unlike the square identities, the final simplified expression has no middle product.

Worked example 9. Calculate 45 × 55 by using their equal distances from 50.

Answer: 45 × 55 = (50 − 5)(50 + 5) = 50² − 5² = 2500 − 25 = 2475. The same two numbers, 50 and 5, appear in both brackets.

How can the identity help find a square?

Rearranging gives a² = (a + b)(a − b) + b². Choose b so that one factor becomes convenient. The added square corrects the difference between the product and the required square.

Worked example 10. Find 197² using a = 197 and b = 3.

Answer: 197² = (197 + 3)(197 − 3) + 3² = 200 × 194 + 9 = 38809. Adding 9 is necessary because the product alone is 197² − 3².

Likewise, 31² = (31 + 1)(31 − 1) + 1² = 32 × 30 + 1 = 961. Both examples use an identity to reorganise the calculation without changing its value.

Before choosing this identity, check the structure of the brackets. They must contain the same two terms with opposite signs between them. Two arbitrary brackets require the general distributive method instead.

How do identities explain numerical patterns?

A pattern suggests a relationship through repeated examples. Algebra can establish why it holds generally. A numerical check verifies a chosen case, whereas an identity explains the relationship for all allowed values of the letters.

What happens when the two square identities are added?

Add (a + b)² = a² + 2ab + b² and (a − b)² = a² − 2ab + b². The opposite middle terms cancel, while the other squares occur twice.

2(a² + b²) = (a + b)² + (a − b)².

Thus twice the sum of two squares can be expressed as the sum of the square of the sum and the square of the difference. For the numbers 5 and 6, this is 2(5² + 6²) = (6 + 5)² + (6 − 5)².

What do three consecutive numbers reveal?

Consecutive numbers follow one another with a difference of 1. Let n be the middle integer. The three numbers are n − 1, n and n + 1. Their outside product is (n − 1)(n + 1) = n² − 1.

Subtracting this product from the middle square gives n² − (n − 1)(n + 1) = 1. The result does not depend on which integer n represents; the variable terms cancel.

Why do calendar diagonal products differ by 7?

For two adjacent dates and the two dates directly below them in a seven-column calendar, let the top-left entry be a. The entries by rows are a, a + 1 and a + 7, a + 8.

The diagonal products multiply opposite corners. Their difference is (a + 1)(a + 7) − a(a + 8) = a² + 8a + 7 − a² − 8a = 7. The example 5 × 11 − 4 × 12 = 55 − 48 illustrates it.

Often, there are multiple ways to solve a problem and arrive at the same correct answer. Expanding different expressions provides a way to check that apparently different methods describe the same relationship.

How can common expansion mistakes be found and corrected?

Checking algebra means checking the operation at each step. Multiplication outside a bracket must reach every term inside. Combining terms is allowed only when their variable parts match. A square applies to the whole expression inside its brackets.

Is every term multiplied correctly?

A constant term contains no variable. Its value still participates in multiplication. An outside factor multiplies the constant term just as it multiplies a term containing a variable.

Consider 2(x − 1) + 3(x + 4), where x is a number. Distributing gives 2x − 2 + 3x + 12 = 5x + 10. Writing 2x − 1 + 3x + 4 incorrectly leaves both constant terms unmultiplied.

Let p and q represent numbers. In −3p(−5p + 2q), multiply rather than add: (−3p)(−5p) + (−3p)(2q) = 15p² − 6pq. The first product has a positive coefficient and the second has a negative coefficient.

Are unlike terms being combined?

Let w represent a number. The terms in 5w² + 6w have different exponents, so the expression does not simplify to 11w². Likewise, a²b and ab² cannot be combined as if their variable parts were identical.

For numbers y, m and n, the expression y + 2(y + 2) simplifies to 3y + 4; it is not (y + 2)². Also, (5m + 6n)² requires the middle term and expands to 25m² + 60mn + 36n².

Is the bracket a sum or a product?

In 3a(2b × 3c), the letters a, b and c represent numbers and the bracket contains a product. Multiplying all factors gives 18abc. Distributing 3a separately over factors would count it twice.

A reliable check is to identify the operation first, include the sign of every term, expand all necessary products, and then collect like terms. A familiar-looking bracket is not enough to justify applying an identity.

How can different counting methods produce equivalent expressions?

Equivalent expressions have the same value for every allowed value of their variables. A picture may be divided into groups in different ways, producing different expressions for the same total. Expansion shows whether the expressions agree.

How many circles are in a growing arrangement?

Let k be the positive integer giving the step number. One method views the arrangement as a square array with k + 1 circles along each side and one corner circle missing. Its total is (k + 1)² − 1.

The diagrams outline a square with one missing corner, a smaller square with two extra strips, a rectangle with an extra strip, and groups each containing three circles in Step 1, four in Step 2, and so on.

Reference: NCERT Class 8, pages 151 to 152

The first method gives k² + 2k + 1 − 1. The third gives k² + k + k. The fourth gives k² + 2k directly by distributivity. All methods therefore count the same total.

How can a square border be counted?

Let n now represent the positive step number in the square-tile pattern. At Step n, the outer square has side n + 2 tiles and the empty inner square has side n tiles. Subtract their areas to count the border.

The expression is (n + 2)² − n² = 4n + 4. The first three figures contain 8, 12 and 16 tiles. A unit square tile has side one unit and area one square unit. Here the formula counts the tiles in the border.

The border method avoids counting corner tiles twice. In both patterns, define the step number before writing a formula, and check that the expression represents the groups actually present in the picture.

How can areas be found by different algebraic methods?

Different area methods can describe the same region. One method may subtract unwanted shapes from a larger shape; another may use the dimensions of the desired region directly. Their algebraic expressions should simplify to the same result.

What happens when four rectangles surround a square?

Let m and n be the shorter and longer sides of each of four equal rectangles. Arrange them around an inner square inside a larger square of side m + n. Here n is greater than m, so the inner side length n − m is positive.

Tadang's method subtracts the four rectangle areas from the outer square: (m + n)² − 4mn. Yusuf's method uses the inner square directly: (n − m)². Expanding the first gives m² + 2mn + n² − 4mn = m² − 2mn + n².

The second expression expands to n² − 2mn + m², which is the same result. This establishes (m + n)² − 4mn = (n − m)² for the arrangement.

How do three rectangles give another area comparison?

Let x and y be the longer and shorter side lengths of three equal rectangles, with x greater than y. Two lie horizontally above and below the region, while the third stands vertically between them. The remaining region between the horizontal rectangles has slanting shading.

Two horizontal rectangles have a vertical rectangle between them. Each has dimensions x and y. Slanting lines mark the spaces on either side of the vertical rectangle.

Reference: NCERT Class 8, page 153, Fig. 1

Anusha's method takes a square of side x and removes the middle rectangle: x² − xy. Vaishnavi's method starts with the outer rectangle of dimensions x and x + 2y, then removes all three rectangles: x(x + 2y) − 3xy.

Aditya's method combines the two side regions to obtain x(x − y). All three expressions simplify to x² − xy. Checking equivalence explains why different choices of the starting region give the same shaded area.

Glossary

• Algebra — The use of letter-numbers to express general patterns, relationships and mathematical reasoning.
• Variable — A letter-number representing a number whose value can be replaced by a chosen number.
• Expression — A combination of numbers or letter-numbers connected by mathematical operations.
• Term — A part of an expression joined through addition or subtraction, with its sign included.
• Coefficient — The numerical multiplier of a variable part within an algebraic term.
• Product — The result obtained by multiplying two or more numbers or expressions.
• Distributive property — The multiplication rule that expands a product with a sum into a sum of products.
• Identity — A statement of equality between expressions that holds for all allowed values of their variables.
• Like terms — Terms with identical variable parts, including matching exponents, whose coefficients can be combined.
• Square — The product obtained by multiplying a number or expression by itself.
• Equivalent expressions — Expressions that have the same value for every allowed value of their variables.
• Consecutive numbers — Numbers following one another in order, with a difference of one between neighbours.

Common errors and misconceptions

• Misconception: In a(b + c), a multiplies just b. Correct: It multiplies both terms, giving ab + ac. Leaving out ac changes the expression.
• Misconception: Increasing both factors by 1 changes their product by a + b. Correct: The change is a + b + 1; the extra 1 comes from multiplying the two added units.
• Misconception: (a + b)² equals a² + b². Correct: The expansion includes 2ab because each bracket contributes a cross product.
• Misconception: (a − b)² ends with −b². Correct: Its final term is +b², because multiplying −b by −b gives b².
• Misconception: 5w² + 6w simplifies to 11w². Correct: The terms have different exponents of w and must remain separate.
• Misconception: Multiplication distributes over the factors in 3a(2b × 3c). Correct: The bracket contains multiplication, so multiply the factors directly to obtain 18abc.
• Misconception: Different-looking counting formulas necessarily give different totals. Correct: Expanding them may produce the same expression, as with the four circle-counting methods.

Exam-style questions with model answers

Q1. Starting with 23 × 27, increase 23 by 1 and decrease 27 by 1. Use distributivity to find the change in the product. [2 marks]

1. The new product is (23 + 1)(27 − 1) = 23 × 27 + 27 − 23 − 1.
2. Subtract the original product. The change is 27 − 23 − 1 = 3, so the product increases by 3.

Q2. Let a and b be two original numbers. Decrease a by 2 and increase b by 3. Expand the new product and express its change from ab. [3 marks]

1. The new factors are a − 2 and b + 3, so their product is (a − 2)(b + 3).
2. Distribute each term: (a − 2)b + 3(a − 2) = ab − 2b + 3a − 6.
3. The original product is ab. Subtracting it from the new expression leaves 3a − 2b − 6, the required change.

Q3. For numbers a and b, expand (3a/2)(a − b + 1/5). Explain whether any terms of the result can be combined. [3 marks]

1. Distribute 3a/2 to all three terms, including the negative term: (3a/2)a − (3a/2)b + (3a/2)(1/5).
2. Simplifying the three products gives (3/2)a² − (3/2)ab + (3/10)a. The fractional coefficients come from multiplying the numerical factors.
3. No terms can be combined further. Their variable parts are a², ab and a, which are different, so these are not like terms.

Q4. A square has side 65 units. Split each side into 60 units and 5 units. Use the areas of the resulting four regions to find the whole area and explain the middle term. [4 marks]

1. The split creates two squares, with sides 60 and 5 units, and two rectangles, each measuring 60 units by 5 units.
2. The square areas are 60² = 3600 and 5² = 25 square units.
3. Each rectangle has area 60 × 5 = 300 square units. Together they contribute 600 square units, corresponding to the middle term 2 × 60 × 5.
4. Adding all four areas gives 3600 + 600 + 25 = 4225 square units, which is the area of the original square.

Q5. At Step k, where k is a positive integer, a circle arrangement is a square array with k + 1 circles in each row and column, with one corner circle removed. Show that (k + 1)² − 1, k² + 2k, k(k + 1) + k and k(k + 2) all give its total, then find the Step 15 total. [5 marks]

1. The complete square array contains (k + 1)² circles. Removing its one missing corner gives the first expression, (k + 1)² − 1.
2. Expanding that expression gives k² + 2k + 1 − 1 = k² + 2k, establishing the second form of the count.
3. For the third expression, distribute k across the bracket: k(k + 1) + k = k² + k + k = k² + 2k.
4. The fourth expression also expands by distributivity: k(k + 2) = k² + 2k. Therefore all four formulas are equivalent counts.
5. At Step 15, substitute k = 15. The count is 15² + 2 × 15 = 225 + 30 = 255 circles.

Q6. Three equal rectangles have longer side x units and shorter side y units, with x greater than y. Two are horizontal, above and below a square of side x, and the third occupies a vertical strip inside that square. The outer rectangle measures x by x + 2y units. Find the area remaining inside the square by two subtraction methods, show their equivalence to x(x − y), and evaluate it for x = 8 and y = 3. [5 marks]

1. The central square has area x² square units. The vertical rectangle inside it has area xy square units, so the first subtraction method gives x² − xy.
2. The outer rectangle has area x(x + 2y) square units. Removing the three equal rectangles leaves the same required region, giving x(x + 2y) − 3xy.
3. Expand the second expression and combine like terms: x² + 2xy − 3xy = x² − xy. Thus both subtraction methods agree.
4. Distributing in x(x − y) gives x² − xy as well. Hence all three expressions describe the same area in different forms.
5. Substitute the given lengths x = 8 and y = 3. The required area is 8(8 − 3) = 8 × 5 = 40 square units.

Q7. A two-by-two calendar block has entries a and a + 1 in its top row, and a + 7 and a + 8 below them, where a is the top-left date. Show that (a + 1)(a + 7) exceeds a(a + 8) by 7. [3 marks]

1. Expand the product of the top-right and bottom-left entries: (a + 1)(a + 7) = a² + 7a + a + 7 = a² + 8a + 7.
2. The other diagonal product is a(a + 8) = a² + 8a, from multiplying the top-left and bottom-right entries.
3. Subtract the second expression from the first. The terms a² and 8a cancel, leaving 7. This proves the stated difference for the given block.

Q8. Calculate 45 × 55 using the difference-of-squares identity. Identify the two numbers used in the identity and explain the cancellation of the middle terms. [3 marks]

1. Write 45 = 50 − 5 and 55 = 50 + 5. Thus the two numbers used in the identity are 50 and 5.
2. In (50 − 5)(50 + 5), the middle products are 50 × 5 and −5 × 50. They cancel because they have equal sizes and opposite signs.
3. The remaining calculation is 50² − 5² = 2500 − 25 = 2475, which is the required product.

Key takeaways

• Distributivity multiplies every term inside a bracket by the outside factor, retaining each term's sign.
• When both factors change, include the product of the changes as well as each change multiplied by an original factor.
• Combine like terms only after checking that their variable parts, including their exponents, match exactly.
• The square of a sum includes two equal cross products, which combine into the middle term 2ab.
• The square of a difference subtracts the middle product 2ab and adds the final square b².
• The difference-of-squares identity works because its two middle products cancel, leaving the two squared terms.
• Quick multiplication methods follow from distributivity together with place value, including correct alignment and carrying.
• Different counting and area methods can lead to equivalent expressions; expanding them explains why their answers agree.

Test yourself

What does a(b + c) mean, and how is it expanded, when a, b and c represent numbers?

It means a multiplied by the entire sum b + c. Distributing a gives ab + ac.

If both factors a and b increase by 1, what is the change from the original product ab?

The change is a + b + 1, since (a + 1)(b + 1) = ab + a + b + 1.

Why can ab and ba be combined, but a²b and ab² cannot generally be combined?

The commutative property makes ab and ba identical variable parts. In a²b and ab², different letters are squared, so they are unlike terms.

What is the expansion of (6x + 5)², where x represents a number?

The expansion is 36x² + 60x + 25. The middle term comes from twice the product of 6x and 5.

Why does subtracting two overlapping strips require adding their overlap back once?

The overlap was subtracted twice, once with each strip. Adding it back once leaves it removed only once.

For numbers a and b, are (a − b)² and (b − a)² equal?

Yes. Expanding either expression gives a² − 2ab + b², so the squared differences are equal.

For an integer n, what is n² − (n − 1)(n + 1)?

The result is 1 because the outside product equals n² − 1, which is one less than the middle square.

A square-tile border has outer side n + 2 tiles and inner empty side n tiles. What is its tile count?

Subtract the inner square from the outer square: (n + 2)² − n² = 4n + 4 tiles.