We Distribute, Yet Things Multiply · Lesson 2 of 5
Special Cases of the Distributive Property
“Derive and apply the square-of-a-sum, square-of-a-difference and difference-of-squares identities.”
• Derive the square-of-a-sum and square-of-a-difference identities. • Explain the middle term using areas and multiplication. • Prove and use the two pattern relationships. • Choose identities for mental calculations and algebraic expansions. • Explain when a geometric picture applies and why the algebra works more widely.
Suppose you know the area of a square with side 60 units and the area of a square with side 5 units. Could you use those two facts to find the area of a square with side 65 units? Adding the two areas seems tempting, but a drawing shows that this leaves two rectangles uncounted. Those missing rectangles explain one of the most common mistakes in algebra and lead to a useful calculation method.
You already know how to multiply every term in one bracket by every term in another. Here you will examine products whose brackets have a special relationship: two identical sums, two identical differences, or a sum paired with a difference. These products occur often enough to deserve familiar forms.
You will derive each form before using it. If you forget a formula, you can recover it by multiplying the brackets. A diagram offers another way to understand the same relationship, while algebra lets you apply it to negative numbers and fractions as well.
Square of the Sum/Difference of Two Numbers
Take a square with side 65 units and split each side into 60 and 5. Draw the lines through the split points. You get one 60 × 60 square, two 60 × 5 rectangles and one 5 × 5 square. The areas are 3600, 300, 300 and 25 square units.
Adding all four areas gives 4225 square units. Adding only 60² and 5² would give 3625, which misses the two rectangles with total area 600. Increasing a side by 5 adds a strip along each of two sides as well as the corner square.
Replace 60 and 5 by a and b. The full square has side a + b, so its area is (a + b)². The two rectangles each have area ab. Their combined area is 2ab; the 2 counts two equal rectangles, rather than squaring ab.
The algebra follows the picture: (a + b)(a + b) = a² + ab + ba + b². Since ab = ba, the middle terms combine. This proves the identity directly from distribution.
Problem
Find 104².
- 1.Choose 104 = 100 + 4, so a = 100 and b = 4.
- 2.The three contributions are a² = 10000, 2ab = 2 × 100 × 4 = 800, and b² = 16.
- 3.Therefore 104² = 10000 + 800 + 16 = 10816. The middle contribution is much larger than 4², so leaving it out would make a substantial error.
Problem
Find 37² by splitting 37 as 30 + 7. Then check 65² using 30 + 35.
- 1.For 37²: 30² + 2 × 30 × 7 + 7² = 900 + 420 + 49 = 1369.
- 2.For 65²: 30² + 2 × 30 × 35 + 35² = 900 + 2100 + 1225 = 4225.
- 3.The split of a side changes the sizes of the four regions, but not the area of the complete square. Some splits make the arithmetic easier than others.
Problem
Expand (6x + 5)² and (3j + 2k)².
- 1.In the first expression, the two parts are 6x and 5. Squaring 6x gives (6x)(6x) = 36x², not 6x².
- 2.The middle term is 2 × 6x × 5 = 60x and the final square is 25. Thus (6x + 5)² = 36x² + 60x + 25.
- 3.For the second, (3j)² = 9j², 2 × 3j × 2k = 12jk, and (2k)² = 4k². The result is 9j² + 12jk + 4k².
- 4.Expanding (3j + 2k)(3j + 2k) independently gives 9j² + 6jk + 6jk + 4k², confirming the same middle term.
Compare (a + b)² with a² + b² when a and b are integers. Try a = 3, b = 2; a = 3, b = −2; and a = 3, b = 0. Explain the general rule.
The difference is 2ab. For 3 and 2 it is 12, so the square of the sum is larger. For 3 and −2 it is −12, so it is smaller. For 3 and 0 it is zero, so the expressions are equal. In general, the square of the sum is larger when ab > 0, smaller when ab < 0, and equal when ab = 0. Two negative values also have a positive product.
Now consider 55² = (60 − 5)². Start with a 60 × 60 square and remove a strip 5 units wide from the right and another from the bottom. Each full strip has area 60 × 5 = 300. However, the strips overlap in the 5 × 5 corner.
Subtracting both complete strips subtracts the corner twice. It should be removed only once, so add back one copy of its area. The remaining area is 3600 − 300 − 300 + 25 = 3025. Adding back the overlap does not put a corner into the remaining picture; it corrects the double subtraction in the calculation.
For general positive lengths with a greater than b, the same argument gives (a − b)² = a² − 2ab + b². You can also derive it without a picture: (a − b)(a − b) = a² − ab − ba + b².
A third route is to write a − b as a + (−b). In the sum identity replace b by −b. The middle term becomes −2ab, but the last term is (−b)² = b². This is why the last square remains positive.
Problem
Find 99² and 58².
- 1.For 99², use (100 − 1)² = 10000 − 2 × 100 × 1 + 1 = 9801.
- 2.For 58², use (60 − 2)² = 3600 − 2 × 60 × 2 + 4.
- 3.Thus 58² = 3600 − 240 + 4 = 3364. In each calculation both strips are subtracted and the small overlap is added back.
Problem
Expand (−2a + 3)².
- 1.Reorder the sum as (3 − 2a)². The two parts are 3 and 2a.
- 2.Apply the difference identity: 3² − 2 × 3 × 2a + (2a)² = 9 − 12a + 4a².
- 3.In descending powers, this is 4a² − 12a + 9. Directly squaring −2a gives +4a², while the mixed product is negative.
- 4.At a = 1, the original is (−2 + 3)² = 1 and the expansion gives 4 − 12 + 9 = 1. This checks this value; the expansion proves the general equality.
Problem
Expand (7y − 3z/4)².
- 1.Take the two parts as 7y and 3z/4. Their squares are 49y² and 9z²/16.
- 2.The middle term is −2 × 7y × 3z/4 = −42yz/4 = −21yz/2.
- 3.The expansion is 49y² − 21yz/2 + 9z²/16. Squaring a fraction squares both its numerator and its denominator.
- 4.Distributing (7y − 3z/4)(7y − 3z/4) produces two terms of −21yz/4, whose sum is the same middle term.
Which is greater, (a − b)² or (b − a)²? Also expand (m + 3)², (6 + p)² and (b − 6)².
The two squared differences are equal: b − a = −(a − b), and opposite numbers have the same square. Their expansions both equal a² − 2ab + b². The other expansions are m² + 6m + 9, p² + 12p + 36, and b² − 12b + 36. Each middle term is twice the product of the two parts, with the appropriate sign.
Investigating Pattern 1
Choose 6 and 5. Twice the sum of their squares is 2(36 + 25) = 122. Now square their sum and their difference: 11² + 1² = 121 + 1 = 122. The same total has appeared in two forms.
Try 5 and 3: twice the sum of their squares is 2(25 + 9) = 68, while 8² + 2² = 64 + 4 = 68. These examples suggest a general pattern. To know that the pattern always works, expand both squared brackets and see which terms cancel.
The sum identity contributes a² + 2ab + b². The difference identity contributes a² − 2ab + b². Adding them cancels +2ab with −2ab. What remains is two copies of a² and two copies of b².
Problem
Write 2(8² + 3²) as the sum of two squares.
- 1.Use the sum 8 + 3 = 11 and the difference 8 − 3 = 5.
- 2.The identity gives 2(8² + 3²) = 11² + 5².
- 3.Check the arithmetic: 2(64 + 9) = 146, and 121 + 25 = 146. The identity explains why the two calculations agree.
Problem
Verify the pattern for a = 3/2 and b = 1/2.
- 1.The sum and difference are 2 and 1, so (a + b)² + (a − b)² = 4 + 1 = 5.
- 2.The other side is 2(9/4 + 1/4) = 2 × 10/4 = 5.
- 3.Both sides agree. More generally, the proof used distribution and cancellation, which are also valid for fractions; the pattern is not restricted to counting numbers.
Pattern 2
Compare 9² − 1² with 10 × 8. Both equal 80. Also, 8² − 6² = 64 − 36 = 28, which equals 14 × 2. The two factors in each product are the sum and difference of the numbers being squared.
Expand (a + b)(a − b): the four products are a², −ab, +ba and −b². The two middle products cancel because ba = ab. The result is a difference of two squares. Notice that this product has different brackets; it is not the square of a difference.
In the drawing, remove a b × b square from an a × a square, with a greater than b. Cut the remaining L-shape into a rectangle a by a − b and a rectangle b by a − b. Rotate and move the smaller piece beside the larger one. The new rectangle has sides a + b and a − b.
The drawing needs positive lengths that fit together. The algebraic identity itself also holds for negative numbers and fractions, because distribution and cancellation remain valid. Do not draw a negative length; use the algebra when the picture no longer represents the values.
Problem
Find 98 × 102 and 45 × 55.
- 1.The first pair is 100 − 2 and 100 + 2. Therefore 98 × 102 = 100² − 2² = 10000 − 4 = 9996.
- 2.The second pair is 50 − 5 and 50 + 5. Therefore 45 × 55 = 50² − 5² = 2500 − 25 = 2475.
- 3.Choose the common centre and the equal distance from it. A product with unequal distances cannot be inserted into this identity in the same way.
Problem
Find 31² and 197² using a² = (a + b)(a − b) + b².
- 1.Rearrange a² − b² = (a + b)(a − b) by adding b² to both sides.
- 2.For 31, choose b = 1 so a − b is 30: 31² = 32 × 30 + 1² = 960 + 1 = 961.
- 3.For 197, choose b = 3 so a + b is 200: 197² = 200 × 194 + 3² = 38800 + 9 = 38809.
- 4.The chosen value of b makes one factor convenient. The added b² restores the amount removed by the difference-of-squares product.
Problem
Express 100 as A² − B² with positive integers A and B.
- 1.We need (A + B)(A − B) = 100. Choose the factor pair 50 and 2.
- 2.Set A + B = 50 and A − B = 2. Adding gives 2A = 52, so A = 26 and B = 24.
- 3.Therefore 100 = 26² − 24² = 676 − 576. The factor pair 10 and 10 instead gives 10² − 0², but that uses zero rather than two positive integers.
A square of a sum has a positive middle term; a square of a difference has a negative middle term; a product of a sum and its matching difference has cancelling middle terms. Check the whole terms, including coefficients and signs, before substituting into an identity.
Quiz
Why does (a + b)² contain 2ab?
Which is the expansion of (x − 4)²?
What is (a + b)² + (a − b)²?
Which identity is most direct for 203 × 197?
When is (a + b)² smaller than a² + b² for integer a and b?
Answer: The expansion contains both ab and ba.. Multiplying the brackets produces a², ab, ba and b². Since ab = ba, the middle terms total 2ab.
Practice Problems
- Expand (m + 3)² and (b − 6)², showing the middle products.
- Expand (2x − 1/2)² and (3a − 9b)(3a + 9b).
- Use suitable identities to find 406², 72², 145², 1097² and 124².
- Use Pattern 1 to evaluate 17² + 7² without squaring 17 and 7 separately.
- A student claims (a − b)² = a² − b² for all integers. Explain the error, give a counterexample, and identify when the equality does happen.
(m + 3)² = m² + 2 × m × 3 + 9 = m² + 6m + 9. (b − 6)² = b² − 2 × b × 6 + 36 = b² − 12b + 36. Both squares have positive final terms. Their middle terms differ in sign because one bracket adds and the other subtracts.
Key Takeaways
• A square of a sum includes two mixed products: (a + b)² = a² + 2ab + b². • A square of a difference is a² − 2ab + b²; adding back an overlap explains its positive last term. • Adding the squared sum and squared difference cancels the mixed terms and gives 2(a² + b²). • Matching sum-and-difference factors give a² − b² because their mixed terms cancel. • Choose a convenient split or common centre to simplify mental calculations. • The algebra applies to negative numbers and fractions even when a positive-length diagram cannot represent them.