Class 8 › Mathematics › Chapter 6
We Distribute, Yet Things Multiply
The distributive property, multiplying algebraic expressions, three standard identities, fast multiplication tricks, and using algebra to explain patterns.
1. Some Properties of Multiplication: Increments in Products
Distributive property of multiplication over addition
$$a(b + c) = ab + ac \qquad (a + b)c = ac + bc$$
The second form follows by commutativity: $(a+b)c = c(a+b) = ca + cb = ac + bc$. Picture $a$ rows with $b$ columns and $c$ columns of dots: $a(b+c)$ total dots $= ab + ac$.
Take $23 \times 27$ and see how the product changes (general $ab$):
| Change | New product | Increase |
|---|---|---|
| $b$ increased by $1$ | $a(b+1) = ab + a$ | $a$ (e.g. $23$) |
| $a$ increased by $1$ | $(a+1)b = ab + b$ | $b$ (e.g. $27$) |
| Both increased by $1$ | $(a+1)(b+1) = ab + a + b + 1$ | $a + b + 1$ (e.g. $23 + 27 + 1 = 51$) |
| $a$ up by $1$, $b$ down by $1$ | $(a+1)(b-1) = ab + b - a - 1$ | $b - a - 1$ (may be negative) |
- Derivation of the both-increase case: treat $(a+1)$ as one term: $(a+1)(b+1) = (a+1)b + (a+1)\cdot 1 = ab + b + a + 1$. You can equally treat $(b+1)$ as one term.
- The product does not always increase: for $a = 10$, $b = 3$: $ab = 30$ but $(a-1)(b-1) = 9 \times 2 = 18$. When $a$ is up $1$ and $b$ is down $1$, the product decreases whenever $a + 1 > b$ (increase $= b - a - 1 < 0$).
- Negative integers: integers also obey the distributive property, so these expressions hold for negative values of $a, b$ too.
Identity: a statement that two algebraic expressions are equal for all values of the letter-numbers (e.g. $a(b+8) = ab + 8a$ and $(a+1)(b-1) = ab + b - a - 1$). Unlike an equation, it is true for every substitution. $\;\ $ Remember: $a(b+c)$ means $a \times (b+c)$; we skip the $\times$ next to brackets.
2. Identity 1: Product of Two Two-Term Expressions
Identity 1
$$(a + m)(b + n) = ab + mb + an + mn$$
Multiply each term of one bracket by each term of the other and add. The increase in $ab$ is $an + bm + mn$. In general $(a+b)(c+d) = ac + ad + bc + bd$.
Area model for $(a+b)^2$ (a special case of Identity 1 with the two brackets equal).
- Decreases: write $b - v$ as $b + (-v)$. Then $(a + u)(b - v) = ab + ub - av - uv$ (take $m = u$, $n = -v$). The signs of terms follow integer multiplication rules.
- $(a - u)(b + v) = ab - ub + av - uv$ and $(a - u)(b - v) = ab - ub - av + uv$.
- Effects: $(a-2)(b+3) = ab + 3a - 2b - 6$ (one number down 2, the other up 3). $(a-3)(b-4) = ab - 4a - 3b + 12$ (both down).
- Unchanged product: $(a+2)(b-4) = ab$ gives $2b - 4a - 8 = 0$, so $b = 2a + 4$. Examples: $(a, b) = (1, 6), (2, 8), (3, 10)$. Check: $(1+2)(6-4) = 3 \times 2 = 6 = 1 \times 6$.
- Multiplication-table frame: if the middle of a $3 \times 3$ frame is $pq$, the others are $(p-1)(q-1),\ (p-1)q,\ (p-1)(q+1),\ p(q-1),\ p(q+1),\ (p+1)(q-1),\ (p+1)q,\ (p+1)(q+1)$.
- Expand (exercise answers): $(3+u)(v-3) = 3v - 9 + uv - 3u$; $\ \tfrac{2}{3}(15 + 6a) = 10 + 4a$; $\ (10a+b)(10c+d) = 100ac + 10ad + 10bc + bd$; $\ (3-x)(x-6) = -x^2 + 9x - 18$; $\ (-5a+b)(c+d) = -5ac - 5ad + bc + bd$; $\ (5+z)(y+9) = 5y + yz + 9z + 45$.
3. Expanding Longer Expressions
The distributive property works with any number of terms inside a bracket. Only like terms (same letter-numbers with the same powers) can be combined.
Worked examples
Example 1: $\dfrac{3a}{2}\left(a - b + \dfrac{1}{5}\right) = \dfrac{3}{2}a^2 - \dfrac{3}{2}ab + \dfrac{3}{10}a$. No two terms are like terms, so no further simplification.
Example 2: $(a+b)(a+b) = a\cdot a + ba + ab + b\cdot b = a^2 + 2ab + b^2$ (since $ba = ab$, they are like terms).
Example 3: $(a+b)(a^2 + 2ab + b^2) = a^3 + a^2b + 2a^2b + 2ab^2 + ab^2 + b^3 = a^3 + 3a^2b + 3ab^2 + b^3$.
Longer products: $(a + ab - 3b^2)(4 + b) = 4a + 5ab + ab^2 - 12b^2 - 3b^3$. $\ (4y+7)(y + 11z - 3) = 4y^2 + 44yz + 77z - 5y - 21$.
Pattern for $(a - b)$ times a growing sum
$$(a-b)(a+b) = a^2 - b^2$$ $$(a-b)(a^2 + ab + b^2) = a^3 - b^3$$ $$(a-b)(a^3 + a^2b + ab^2 + b^3) = a^4 - b^4$$
Next in the pattern: $(a-b)(a^4 + a^3b + a^2b^2 + ab^3 + b^4) = a^5 - b^5$.
- Exponent rules used: $a \times a^2 = a^3$; $ba^2 = a^2b$; $a \times 2ab = 2a^2b$.
- Which is greater, $(a-b)^2$ or $(b-a)^2$? They are equal, since $b - a = -(a-b)$ and a square is the same for a number and its negative.
4. A Pinch of History
- The distributive property was used implicitly in ancient Egypt, Mesopotamia, Greece, China and India. Euclid (geometric form) and Aryabhata (algebraic form) used it often.
- Brahmasphutasiddhanta (Brahmagupta, 628 CE), Verse 12.55: the first explicit statement, calling multiplication by splitting the multiplier khanda-gunanam ("multiplication by parts"): break the multiplier into parts that sum to it, multiply the multiplicand by each part and add. In letters: $(a + b)c = ac + bc$.
- Verse 12.56 gives fast multiplication, called ista-gunana. Sridharacharya (750 CE) and Bhaskaracharya (Lilavati, 1150 CE) also wrote on such methods.
5. Fast Multiplication Tricks
By 11
$N \times 11 = N \times 10 + N$
$3874 \times 11 = 38740 + 3874 = 42614$. In one line: write the units digit, then the sum of each pair of adjacent digits (from the right), carrying as needed, then the first digit. $3874$: $4$, $(7+4) = 11 \to 1$ carry $1$, $(8+7) + 1 = 16 \to 6$ carry $1$, $(3 + 8) + 1 = 12 \to 2$ carry $1$, $3 + 1 = 4$, giving $42614$.
By 101
$N \times 100 + N$
Digits $dcba \times 101$: the digits $d, c$ stay at the front, then $(b + d)$, then $(a + c)$, then $b, a$ at the end (carry when a sum exceeds 9). Example: $3874 \times 101 = 387400 + 3874 = 391274$.
By 1001, 10001
$N \times 10^k + N$
$265831 \times 1001 = 265831000 + 265831 = 266096831$. $\;1111 \times 1001 = 1112111$. $\;dcba \times 10001 = dcbadcba$ (digits repeat).
By 99, 999
$N \times 10^k - N$
$9734 \times 99 = 9734(100 - 1) = 973400 - 9734 = 963666$. $\;23478 \times 999 = 23478000 - 23478 = 23454522$.
- Evaluate: $94 \times 11 = 1034$; $\;495 \times 11 = 5445$; $\;3279 \times 11 = 36069$; $\;4791256 \times 11 = 52703816$; $\;89 \times 101 = 8989$; $\;949 \times 101 = 95849$.
- General rule for 11: every product digit is the sum of two adjacent digits of the number (with the first and last digits copied), then carries are adjusted.
6. Special Cases: Square of a Sum and a Difference
Identity 1A (square of a sum)
$$(a + b)^2 = a^2 + 2ab + b^2$$
Identity 1B (square of a difference)
$$(a - b)^2 = a^2 - 2ab + b^2$$
Geometric meaning
$65^2$: a square of side $65 = 60 + 5$ splits into a $60 \times 60$ square, a $5 \times 5$ square and two $60 \times 5$ rectangles: $65^2 = 3600 + 25 + 600 = 4225$.
$55^2 = (60 - 5)^2$: start with $60^2$, remove two $60 \times 5$ rectangles (removing the corner $5 \times 5$ twice), so add back $5^2$: $3600 - 300 - 300 + 25 = 3025$.
Algebra: $(a-b)^2 = (a-b)(a-b) = a^2 - ab - ba + b^2 = a^2 - 2ab + b^2$. Or $(a - b)^2 = (a + (-b))^2$ and use 1A.
- Mental squares: $104^2 = (100+4)^2 = 10816$; $\;37^2 = (40-3)^2 = 1369$; $\;99^2 = (100-1)^2 = 9801$; $\;58^2 = (60-2)^2 = 3364$; $\;46^2 = (40+6)^2 = 2116$; $\;91^2 = (100-9)^2 = 8281$; $\;406^2 = 164836$; $\;72^2 = 5184$; $\;145^2 = 21025$; $\;1097^2 = 1203409$; $\;124^2 = 15376$.
- With coefficients: $(6x+5)^2 = 36x^2 + 60x + 25$; $\;(3j + 2k)^2 = 9j^2 + 12jk + 4k^2$; $\;(b-6)^2 = b^2 - 12b + 36$; $\;(-2a+3)^2 = 4a^2 - 12a + 9$; $\;(7y - 4z)^2 = 49y^2 - 56yz + 16z^2$; $\;(2x - \tfrac{1}{2})^2 = 4x^2 - 2x + \tfrac{1}{4}$; $\;(6x+5y)^2 = 36x^2 + 60xy + 25y^2$.
- If you forget the identity, just apply the distributive property: $(6x+5)(6x+5)$.
- Is $(a+b)^2 > a^2 + b^2$? $(a+b)^2 = a^2 + b^2 + 2ab$, so it is greater exactly when $ab > 0$.
7. Difference of Squares
Identity 1C
$$(a + b)(a - b) = a^2 - b^2$$
Proof: $(a+b)(a-b) = a^2 - ab + ba - b^2 = a^2 - b^2$ (the middle terms cancel).
Cut the L-shape ($a^2 - b^2$) along the dotted line and move the top piece to the side: it becomes a rectangle $(a+b)$ by $(a-b)$.
- Fast products: $98 \times 102 = (100-2)(100+2) = 9996$; $\;45 \times 55 = (50-5)(50+5) = 2475$; $\;397 \times 403 = (400-3)(400+3) = 159991$; $\;43 \times 45 = (44-1)(44+1) = 1935$.
- Sridharacharya's square method (750 CE): $a^2 = (a+b)(a-b) + b^2$. Choose $b$ so that $a \pm b$ is easy: $31^2 = 32 \times 30 + 1 = 961$; $\;197^2 = 200 \times 194 + 9 = 38809$.
- $100$ as a difference of two squares: $100 = 26^2 - 24^2$ ($676 - 576$). More generally $a^2 - b^2 = 100$ needs $(a+b)(a-b) = 100$ with both factors of the same parity: $50 \times 2$.
- Which is bigger (without full multiplication)? $14 \times 26 = (16-2)(24+2) = 16 \times 24 + 32 - 48 - 4 = 16 \times 24 - 20$, so $16 \times 24$ is larger. $\;25 \times 75 = (26-1)(74+1) = 26 \times 74 + 26 - 74 - 1 = 26 \times 74 - 49$, so $26 \times 74$ is larger.
8. Investigating Patterns
Pattern 1: twice a sum of squares
Examples: $2(2^2 + 1^2) = 3^2 + 1^2$, $\;2(5^2 + 6^2) = (6+5)^2 + (6-5)^2$.
Add Identities 1A and 1B: $(a+b)^2 + (a-b)^2 = 2a^2 + 2b^2$, so $\;2(a^2 + b^2) = (a+b)^2 + (a-b)^2$.
Pattern 2: difference of squares
Examples: $9 \times 9 - 1 \times 1 = 10 \times 8$; $\;8 \times 8 - 6 \times 6 = 14 \times 2$. Both are $a^2 - b^2 = (a+b)(a-b)$, which is Identity 1C.
- Because these are identities, both patterns hold for all numbers: negative integers and fractions as well as counting numbers.
- Three consecutive numbers: $n^2 - (n-1)(n+1) = n^2 - (n^2 - 1) = 1$. The middle number squared minus the product of the other two is always $1$.
- "Add two numbers and multiply by half their sum": $(a+b) \times \tfrac{1}{2}(a+b) = \tfrac{1}{2}(a+b)^2$, which is half of the square of the sum.
- Calendar $2 \times 2$ squares with $a, a+1, a+7, a+8$: diagonal products are $a(a+8) = a^2 + 8a$ and $(a+1)(a+7) = a^2 + 8a + 7$. Their difference is always $7$ (e.g. $3 \times 11 = 33$ and $4 \times 10 = 40$).
9. Mind the Mistake, Mend the Mistake
| # | Given | Verdict | Correct working |
|---|---|---|---|
| 1 | $-3p(-5p + 2q) = -3p + 5p - 2q = p - 2q$ | Mistake | Multiply every term, including the product $p \times p$: $15p^2 - 6pq$ |
| 2 | $2(x-1) + 3(x+4) = 2x - 1 + 3x + 4 = 5x + 3$ | Mistake | The $2$ and $3$ multiply both terms: $2x - 2 + 3x + 12 = 5x + 10$ |
| 3 | $y + 2(y+2) = (y+2)^2 = y^2 + 4y + 4$ | Mistake | $y + 2(y+2)$ is not $(y+2)(y+2)$: $y + 2y + 4 = 3y + 4$ |
| 4 | $(5m + 6n)^2 = 25m^2 + 36n^2$ | Mistake | Missing middle term: $25m^2 + 60mn + 36n^2$ |
| 5 | $(-q + 2)^2 = q^2 - 4q + 4$ | Correct | $q^2 - 2 \cdot q \cdot 2 + 4$ |
| 6 | $3a(2b \times 3c) = 6ab \times 9ac = 54a^2bc$ | Mistake | $3a$ multiplies the whole product once: $3a \times 6bc = 18abc$ |
| 7 | $\tfrac{1}{2}(10s - 6) + 3 = 5s - 3 + 3 = 5s$ | Correct | |
| 8 | $5w^2 + 6w = 11w^2$ | Mistake | $w^2$ and $w$ are not like terms; the sum stays $5w^2 + 6w$ |
| 9 | $2a^3 + 3a^3 + 6a^2b + 6ab^2 = 5a^3 + 12a^2b^2$ | Mistake | Only $2a^3 + 3a^3$ combine: $5a^3 + 6a^2b + 6ab^2$ |
| 10 | $(x+2)(x+5) = x^2 + 2x + 5x + 10 = x^2 + 7x + 10$ | Correct | |
| 11 | $(a+2)(b+4) = ab + 8$ | Mistake | Four products: $ab + 4a + 2b + 8$ |
| 12 | $ab^2 + a^2b + a^2b^2 = ab(a + b + ab)$ | Correct | Factor out $ab$: $ab \cdot b$, $ab \cdot a$, $ab \cdot ab$ |
10. This Way or That Way: Many Ways, Same Answer
The circle pattern with $3, 8, 15, 24, \dots$ circles (Step 1, 2, 3, 4, …) can be seen in four ways:
| Method | Expression for Step $k$ | Simplified |
|---|---|---|
| 1: square minus a unit | $(k+1)^2 - 1$ | $k^2 + 2k$ |
| 2: square plus two columns | $k^2 + 2 \times k$ | $k^2 + 2k$ |
| 3: rectangle plus a column | $k(k+1) + k$ | $k^2 + 2k$ |
| 4: one rectangle | $k(k+2)$ | $k^2 + 2k$ |
- All correct methods give the same expression, $k^2 + 2k$. Step 15 has $15^2 + 30 = 255$ circles.
- Square-tile pattern: Step $n$ has $(n+2)^2 - n^2$ tiles. Step 1: $3^2 - 1^2 = 8$; Step 4: $6^2 - 4^2$; Step 10: $12^2 - 10^2 = 44$. (It simplifies to $4n + 4$.)
- Unit-block patterns (exercise): (a) $(y+2)^2$: Step 10 $= 144$. (b) $(y+1)^2 + y$: Step 1: $5$, Step 2: $11$, Step 3: $19$, Step 4: $29$; Step 10 $= 121 + 10 = 131$.
- Inner shaded square (four rectangles in a square): Tadang: $(m+n)^2 - 4mn$. Yusuf: $(n-m)^2$. Expanding both gives $m^2 + n^2 - 2mn$, so they agree.
- Hatched region ($x$, $y$): $x^2 - xy$ (Anusha), $x(x+2y) - 3xy$ (Vaishnavi) and $x(x-y)$ (Aditya) are all equal to $x^2 - xy$. For $x = 8$, $y = 3$: $8 \times 5 = 40$.
- Dashed region ($p, r, s$): $ps - pr - sr + r^2$ $= (p - r)(s - r)$. With $p = 6$, $r = 3.5$, $s = 9$: $13.75$ square units.
11. Applications and Exercise Facts
- "Two more than a square number": $s^2 + 2$. "Sum of the squares of two consecutive numbers": $m^2 + (m+1)^2$.
- Remainders on division by 7: $n_1 = 7a + 3$, $n_2 = 7b + 5$. The sum leaves remainder $1$ ($8$); the difference (larger minus smaller, i.e. $n_2 - n_1$) leaves $2$ (the other order leaves $5$); the product leaves $1$ ($15$).
- Park in Dhauli: length $2g + 4w$, breadth $g + 2w$. Area to tile $= (2g + 4w)(g + 2w) - 2g^2 = 8gw + 8w^2 = 8w(w + g)$ square feet.
- Claims (exercise):
(i) $(k+1)(k+2) - (k+3) = k^2 + 2k - 1$, not always $2$: False.
(ii) $(2q+1)(2q-3) = 4q^2 - 4q - 3 = 4(q^2 - q) - 3$, never a multiple of $4$: False.
(iii) Even squares are multiples of $4$; odd squares are $1$ more than a multiple of $8$ [$(2n+1)^2 = 4n(n+1) + 1$ and $n(n+1)$ is even]: True.
(iv) $(6n+2)^2 - (4n+3)^2 = 20n^2 - 5$; adding $5$ gives $20n^2$, which is not a perfect square in general: False.
12. Summary and Quick Revision
| Identity | Formula | Use |
|---|---|---|
| Distributive | $a(b + c) = ab + ac$ | Expanding any bracket |
| Identity 1 | $(a+m)(b+n) = ab + mb + an + mn$ | Change in a product; general expansion |
| Identity 1A | $(a+b)^2 = a^2 + 2ab + b^2$ | Squares like $46^2 = (40+6)^2$ |
| Identity 1B | $(a-b)^2 = a^2 - 2ab + b^2$ | Squares like $99^2 = (100-1)^2$ |
| Identity 1C | $(a+b)(a-b) = a^2 - b^2$ | Products like $98 \times 102$ |
| Derived | $2(a^2 + b^2) = (a+b)^2 + (a-b)^2$; $\ a^2 = (a+b)(a-b) + b^2$ | Patterns; Sridharacharya's squares |
| Cubes | $(a+b)(a^2 + 2ab + b^2) = a^3 + 3a^2b + 3ab^2 + b^3$ | Longer expansions |
Common mistakes:
• Writing $(a+b)^2 = a^2 + b^2$ (the middle term $2ab$ is missing); likewise $(a-b)^2 \neq a^2 - b^2$.
• Forgetting that a number outside brackets multiplies every term: $2(x-1) = 2x - 2$, not $2x - 1$.
• Mishandling signs when expanding $(a - u)(b - v)$: the last term is $+uv$.
• Adding unlike terms: $5w^2 + 6w \neq 11w^2$; $a^2b$ and $ab^2$ are different.
• Treating $y + 2(y+2)$ as $(y+2)^2$.
• Using an identity as if it were a one-off fact: it holds for all numbers, including negatives and fractions.
• Concluding from one example. A single counterexample disproves a claim (e.g. $(k+1)(k+2) - (k+3) = 2$ fails at $k = 1$), but examples never prove it.
• Writing $(a+b)^2 = a^2 + b^2$ (the middle term $2ab$ is missing); likewise $(a-b)^2 \neq a^2 - b^2$.
• Forgetting that a number outside brackets multiplies every term: $2(x-1) = 2x - 2$, not $2x - 1$.
• Mishandling signs when expanding $(a - u)(b - v)$: the last term is $+uv$.
• Adding unlike terms: $5w^2 + 6w \neq 11w^2$; $a^2b$ and $ab^2$ are different.
• Treating $y + 2(y+2)$ as $(y+2)^2$.
• Using an identity as if it were a one-off fact: it holds for all numbers, including negatives and fractions.
• Concluding from one example. A single counterexample disproves a claim (e.g. $(k+1)(k+2) - (k+3) = 2$ fails at $k = 1$), but examples never prove it.