We Distribute, Yet Things Multiply

1. The big idea — distributivity

Algebra lets us write general statements and prove patterns. The single property that ties multiplication and addition together is distributivity. For any three numbers $a,b,c$:

Read it as: "$a$ times the bracket = $a$ times each piece, added up." Here $a(b+c)$ means $a\times(b+c)$ — we usually drop the $\times$ next to brackets, just like in $5a$ or $xy$.

Picture it as a dot grid: $a$ rows of $(b+c)$ dots split into two blocks — one $a\times b$ block ($ab$ dots) and one $a\times c$ block ($ac$ dots). The total is $ab+ac$.

It works the other way too, using commutativity: $(a+b)c = c(a+b)=ca+cb=ac+bc$. And it holds for negative integers and fractions as well, e.g. $x(y+z)=xy+xz$ for any integers $x,y,z$.

Why this chapter's odd name? When we "distribute" the multiplier across a bracket, the single product breaks into many products that are then added — we distribute, yet things multiply. Every shortcut, identity and area-trick in the chapter is just this one law applied cleverly. A pinch of history: the distributive law was used implicitly by Euclid (geometrically) and Āryabhaṭa (algebraically); the first explicit statement came from Brahmagupta in the Brahmasphuṭasiddhānta, who called multiplication-by-parts khaṇḍa-guṇanam.

2. Increase in a product

Take $23\times27$. What happens to the product when we nudge the numbers?

• Second number $+1$: $23(27+1)=23\times27+23$ — product grows by the first number, $23$.
• First number $+1$: by symmetry, $a(b+1)$... actually $(a+1)b=ab+b$ — grows by the other number.
• Both $+1$: $(a+1)(b+1)$ — let's expand it.

Treat $(a+1)$ as one term and distribute:

So when both grow by $1$, the product $ab$ increases by $a+b+1$. Check with numbers: $(23+1)(27+1)=23\times27+(27+23+1)=621+51=672=24\times28.$ Correct.

3. One up, one down — and identities

If $a$ goes up by $1$ and $b$ goes down by $1$:

The product can rise or fall depending on the numbers. Statements like $a(b+8)=ab+8a$ or $(a+1)(b-1)=ab+b-a-1$ are true for all values — such always-true equalities are called identities.

4. Multiplying two brackets — Identity 1

If $a$ increases by $m$ and $b$ increases by $n$, the new product is $(a+m)(b+n)$. Distribute twice:

Golden rule: the product is the sum of (each term of the first bracket) $\times$ (each term of the second bracket). The increase here is $an+bm+mn$.

The same identity handles decreases — just use negative values of $m$ or $n$ and follow the sign rules of integer multiplication:

• $(a+u)(b-v)=ab-uv+... = ab+ub-av-uv$ (take $n=-v$).
• $(a-u)(b+v)=ab-ub+av-uv$.
• $(a-u)(b-v)=ab-ub-av+uv$.

One identity, every case — that is the power of letting letters take negative values. The signs simply follow integer-multiplication rules: a $+$ term times a $-$ term gives a $-$ term, and a $-$ times a $-$ gives a $+$. So you never need to memorise four separate rules; Identity 1 with the right signs does all of them.

Worked check (Identity 1 with $m=-2,\ n=+3$): the product changes by $an+bm+mn = 3a-2b-6.$ For $a=23,\ b=27$: $(23-2)(27+3)=21\times30=630$, and $621+(3\times23-2\times27-6)=621+(69-54-6)=621+9=630.$ Matches.

5. Like terms — when can we combine?

Terms with exactly the same letter-numbers are like terms and only those can be added into one term. So $\tfrac32 a^2$ and $\tfrac{3}{10}a$ cannot be combined, but $ab$ and $ba$ can ($ba=ab$).

6. Fast multiplication using distributivity

Brahmagupta called this khaṇḍa-guṇanam (multiplication by parts) and ista-guṇana. Split one number into easy parts.

• $\times 11$: $3874\times11=3874(10+1)=38740+3874=42614.$ Trick: write the number, add each pair of neighbouring digits (with carries). $94\times11=1034,\ 495\times11=5445.$
• $\times 101$: $3874\times101=3874(100+1)=387400+3874=391274.$ So $89\times101=8989,\ 949\times101=95849.$
• $\times 1001,\ \times 99,\ \times 999$: $23478\times999=23478(1000-1)=23478000-23478=23454522.$ Also $9734\times99=9734(100-1)=973400-9734=963666.$

The general digit-rule for $\times 11$ on a number with digits $\dots dcba$ comes straight from $\overline{dcba}\times(10+1)$: each output digit is the sum of two neighbouring digits (carry when over $9$). For $\times 101$ on $\overline{dcba}$, you add the number to itself shifted two places, giving digit-sums two columns apart. All of it is the distributive property in disguise — no new "rule" to learn.

7. Special Case A — square of a sum

A square of side $65$ splits into a $60\times60$ square, a $5\times5$ square, and two $60\times5$ rectangles. So $65^2=(60+5)^2=60^2+2(60\times5)+5^2=3600+600+25=4225.$ In general:

Note $(a+b)^2$ is not $a^2+b^2$ — the extra $2ab$ is the two rectangles. For positive $a,b$ it is always greater than $a^2+b^2$.

8. Special Case B — square of a difference

A square of side $55$ sits inside a side-$60$ square; remove two $60\times5$ rectangles but add back the $5\times5$ counted twice: $(60-5)^2=60^2-(60\times5)-(5\times60)+5^2=3600-300-300+25=3025.$ In general, since $(a-b)^2=(a+(-b))^2$:

9. Special Case C — product of sum and difference

From the pattern $9\times9-1\times1=10\times8$, $8\times8-6\times6=14\times2$, etc., we spot $a^2-b^2=(a+b)(a-b)$. Expanding confirms it: $(a+b)(a-b)=a^2-ab+ba-b^2=a^2-b^2$ (the $ab$ terms cancel).

Sridharacharya's squaring trick rearranges 1C as $a^2=(a+b)(a-b)+b^2$: e.g. $31^2=(31+1)(31-1)+1^2=32\times30+1=961$, and $197^2=(197+3)(197-3)+3^2=200\times194+9=38809.$

10. Mind the mistake — common errors (§6.3)

NCERT asks you to spot wrong "simplifications". The classic traps:

• $(y+2)^2 \ne y^2+4$ — you dropped $2ab$. Correct: $y^2+4y+4.$
• $(5m+6n)^2 \ne 25m^2+36n^2$ — missing $2\times5m\times6n$. Correct: $25m^2+60mn+36n^2.$
• $(-q+2)^2 = q^2-4q+4$ (treat as $(2-q)^2$), not $q^2-4q+4$ with wrong signs — check carefully.
• $5w^2+6w \ne 11w^2$ — $w^2$ and $w$ are not like terms; they cannot be combined.
• $2(x-1)+3(x+4)=2x-2+3x+12=5x+10$ — don't forget to distribute the $-1$ and the $+12$.

11. Many ways, one answer (§6.4) — patterns & areas

A dot pattern's count at Step $k$ can be written four ways — all equal:

Different correct expressions for the same figure must be identically equal — proven by expanding. Same with areas: an interior square framed by four equal $m\times n$ rectangles has area $(m+n)^2-4mn$ (Tadang) and also $(n-m)^2$ (Yusuf). Expanding both: $(m+n)^2-4mn=m^2+2mn+n^2-4mn=m^2-2mn+n^2=(n-m)^2.$ Equal.

12. Two useful number patterns

• Pattern 1: $2(a^2+b^2)=(a+b)^2+(a-b)^2.$ (Add Identities 1A and 1B: the $2ab$ cancels, $a^2+a^2=2a^2$.) Check: $2(5^2+6^2)=11^2+1^2.$
• Pattern 2: $a^2-b^2=(a+b)(a-b)$ — Identity 1C again, seen as $9\times9-1=10\times8$, etc.

Both patterns hold for negative integers and fractions too, since the identities themselves do.

More gems you can prove with identities:

• Squares of even numbers are multiples of 4: an even number is $2k$, so $(2k)^2=4k^2.$
• Squares of odd numbers are 1 more than a multiple of 8: an odd number is $2k+1$, so $(2k+1)^2=4k^2+4k+1=4k(k+1)+1$; since $k(k+1)$ is always even, $4k(k+1)$ is a multiple of $8$.
• Consecutive-number trick: for three consecutive numbers, $(\text{middle})^2-(\text{first}\times\text{last})=a^2-(a-1)(a+1)=a^2-(a^2-1)=1$ — always exactly $1$.
• Difference of squares of two consecutive numbers: $(n+1)^2-n^2=2n+1$ — always the odd number in between, by Identity 1C.

13. NCERT Exercises — selected solutions

Expand products (Q2, p.143):

• (i) $(3+u)(v-3)=3v-9+uv-3u.$
• (ii) $\dfrac23(15+6a)=10+4a.$
• (iii) $(10a+b)(10c+d)=100ac+10ad+10bc+bd.$
• (iv) $(3-x)(x-6)=3x-18-x^2+6x=-x^2+9x-18.$
• (vi) $(5+z)(y+9)=5y+45+zy+9z.$

Compute with identities (Q1, p.154):

• (i) $46^2=(40+6)^2=1600+480+36=2116$ (Identity 1A).
• (ii) $397\times403=(400-3)(400+3)=160000-9=159991$ (Identity 1C).
• (iii) $91^2=(90+1)^2=8281$ (or 1B as $(100-9)^2$).
• (iv) $43\times45=(44-1)(44+1)=1936-1=1935$ (Identity 1C).

Use an identity or distributivity (Q2, p.154):

• (i) $(p-1)(p+11)=p^2+10p-11.$
• (ii) $(3a-9b)(3a+9b)=9a^2-81b^2$ (1C).
• (iv) $(6x+5y)^2=36x^2+60xy+25y^2$ (1A).
• (v) $\left(2x-\dfrac12\right)^2=4x^2-2x+\dfrac14$ (1B).
• (vi) $(7p)(3r)(p+2)=21pr(p+2)=21p^2r+42pr.$

Calendar 2×2 box (Q4, p.155): label the box $a,\ a+1,\ a+7,\ a+8.$ Diagonal products: $a(a+8)=a^2+8a$ and $(a+1)(a+7)=a^2+8a+7.$ The two diagonal products always differ by exactly $7$ — that is why $5\times11=55$ and $4\times12=48$ differ by $7$.

14. Quick revision checklist

• Distributive: $a(b+c)=ab+ac$ — multiply into every term.
• Two brackets: each term $\times$ each term $\Rightarrow (a+m)(b+n)=ab+mb+an+mn.$
• $(a+b)^2=a^2+2ab+b^2$;   $(a-b)^2=a^2-2ab+b^2$;   $(a+b)(a-b)=a^2-b^2.$
• Never drop the middle $2ab$; never combine unlike terms ($w^2\ne$ part of $w$).
• Use $(a\pm b)$ splits for fast squares and near-round products.