We Distribute Yet Things Multiply

Chapter at a Glance

This chapter details the algebraic and geometric properties of the distributive property of multiplication over addition, showing how algebraic expressions are expanded, simplified, and used to describe patterns. By representing multiplication as areas of partitioned rectangles, students learn to expand products of binomials, simplify like terms, and identify algebraic identities. The chapter also details mental math shortcuts (like multiplying by $11$, $101$, or $99$), reviews binomial square expansions, and highlights historical contributions from Indian mathematicians like Brahmagupta and Sridharacharya.

Key Definitions & Terminology

• Distributive Property: The rule stating that multiplying a sum by a number is equivalent to multiplying each addend individually and adding the products: $a(b+c) = ab + ac$.
• Identity: An algebraic equation that remains true for any numerical values substituted for its variables.
• Like Terms: Terms in an algebraic expression that share the exact same variable factors raised to the same exponents (e.g., $5a^2b$ and $-2a^2b$).
• Khanda-gunanam: The Sanskrit term used by Brahmagupta (628 CE) meaning "multiplication by parts," referring to the distributive property.
• Ista-gunana: The historical Indian term for rapid calculation techniques using the distributive property.
• Binomial: An algebraic expression containing exactly two terms (e.g., $x+y$).

Formulas, Rules & Properties

• The Binomial Product Formula:
  $$(a+b)(c+d) = ac + ad + bc + bd$$
• Sign Variations of Binomial Products:
  $$(a+u)(b-v) = ab - av + ub - uv$$
  $$(a-u)(b+v) = ab + av - ub - uv$$
  $$(a-u)(b-v) = ab - av - ub + uv$$
• Standard Algebraic Identities:
• Identity 1A (Square of a Sum):
  $$(a+b)^2 = a^2 + 2ab + b^2$$
• Identity 1B (Square of a Difference):
  $$(a-b)^2 = a^2 - 2ab + b^2$$
• Identity 1C (Difference of Squares):
  $$(a+b)(a-b) = a^2 - b^2$$
• Parity and Squaring Patterns:
• Sum of Squares Identity:
  $$2(a^2 + b^2) = (a+b)^2 + (a-b)^2$$
• Difference of Powers Factorization:
  $$(a-b)(a^{n-1} + a^{n-2}b + \dots + ab^{n-2} + b^{n-1}) = a^n - b^n$$
• Sridharacharya's Squaring Identity:
  $$a^2 = (a+b)(a-b) + b^2$$

Core Concepts & Topics

• Geometric Area Visualizations:
• Square of a Sum: A square with sidelength $a+b$ has a total area of $(a+b)^2$. By partitioning it, it is divided into a square of area $a^2$, a square of area $b^2$, and two rectangles of area $ab$, illustrating $(a+b)^2 = a^2 + 2ab + b^2$.
• Square of a Difference: A square with sidelength $a-b$ is constructed inside a larger square of side $a$. Its area is computed by taking the total area $a^2$, subtracting two overlapping rectangular strips of size $a \times b$, and then adding back the corner square of side $b$ (area $b^2$) that was subtracted twice: $(a-b)^2 = a^2 - 2ab + b^2$.
• Brahmagupta's Khaṇḍa-Guṇanam:
• Defined in the Brahmasphuṭasiddhānta (Verse 12.55): "The multiplier is broken up into two or more parts whose sum is equal to it; the multiplicand is then multiplied by each of these and the results added."
• Calendar Diagonal Constants:
• In a monthly calendar page, any $2 \times 2$ block of dates can be modeled as:
  $$\begin{pmatrix} x & x+1 \ x+7 & x+8 \end{pmatrix}$$
  The diagonal products are:
  $$\text{Product}_1 = x(x+8) = x^2 + 8x$$
  $$\text{Product}_2 = (x+1)(x+7) = x^2 + 8x + 7$$
  The difference between these diagonal products is always a constant:
  $$\text{Product}_2 - \text{Product}_1 = (x^2+8x+7) - (x^2+8x) = 7$$

Worked Examples

• Fast Arithmetic with 11 and 101:
• Problem: Compute $3874 \times 11$ and $9734 \times 99$ using the distributive property.
• Solution:
  • For $3874 \times 11$: Write the digits from right to left, adding adjacent terms:
  • Units place: $4$
  • Tens place: $4+7 = 11 \rightarrow 1$ (carry $1$)
  • Hundreds place: $7+8 = 15 + 1 \text{ (carry)} = 16 \rightarrow 6$ (carry $1$)
  • Thousands place: $8+3 = 11 + 1 \text{ (carry)} = 12 \rightarrow 2$ (carry $1$)
  • Ten-thousands place: $3 + 1 \text{ (carry)} = 4$
  • Result: $42,614$.
  • For $9734 \times 99$: Expand as $9734 \times (100 - 1)$:
    $$9734 \times 100 - 9734 = 973,400 - 9,734 = 963,666$$
• Finding Squares using Sridharacharya's Method (Page 196):
• Problem: Find $31^2$ and $197^2$ using the formula $a^2 = (a+b)(a-b) + b^2$.
• Solution:
  • For $31^2$: Let $a=31$, choose $b=1$:
    $$31^2 = (31+1)(31-1) + 1^2 = 32 \times 30 + 1 = 960 + 1 = 961$$
  • For $197^2$: Let $a=197$, choose $b=3$ to round to $200$:
    $$197^2 = (197+3)(197-3) + 3^2 = 200 \times 194 + 9 = 38,800 + 9 = 38,809$$
• Expressing 100 as the Difference of Squares (Figure it Out Q2, Page 196):
• Problem: Express $100$ as $x^2 - y^2$.
• Solution: Write $x^2 - y^2 = (x-y)(x+y) = 100$.
  • Choose factor pairs of $100$ of the same parity: $x-y = 2$ and $x+y = 50$.
  • Adding the equations: $2x = 52 \rightarrow x = 26$.
  • Subtracting the equations: $2y = 48 \rightarrow y = 24$.
  • Verify: $26^2 - 24^2 = 676 - 576 = 100$.
• Shaded Area Equivalence (Page 200):
• Problem: Verify the equivalence of these three formulas for the same shaded area: Anusha ($x^2 - xy$), Vaishnavi ($x(x+2y) - 3xy$), and Aditya ($2x(\frac{x-y}{2})$).
• Solution:
  • Anusha: $x^2 - xy$
  • Vaishnavi: $x(x+2y) - 3xy = x^2 + 2xy - 3xy = x^2 - xy$
  • Aditya: $2x(\frac{x-y}{2}) = x(x-y) = x^2 - xy$
  • All three methods are equivalent. For $x=8, y=3$, the area is $8^2 - 8(3) = 64 - 24 = 40$.
• Dhauli Park Path Area (Page 203 Q10):
• Problem: A park has two green square plots of area $g^2$ sq. ft. A walking path of width $w$ surrounds and separates them. Write an expression for the path area.
• Solution:
  • Total length of the park layout $= (2g + 4w)$ ft.
  • Total width of the park layout $= (g + 2w)$ ft.
  • Total area $= (2g + 4w)(g + 2w) = 2g^2 + 8gw + 8w^2$ sq. ft.
  • Area of the two green squares $= 2g^2$ sq. ft.
  • Area of the path $= \text{Total area} - \text{Green area} = 8gw + 8w^2 = 8w(g+w)$ sq. ft.

Practical Activities & Experiments

• Visual Algebraic Puzzle: Cut a $10\text{ cm} \times 10\text{ cm}$ square of grid paper. Mark it to form a $7\text{ cm}$ square, a $3\text{ cm}$ square, and two $7\text{ cm} \times 3\text{ cm}$ rectangles. Color them differently, cut them out, and assemble them to demonstrate to others that $(7+3)^2 = 7^2 + 2(7 \times 3) + 3^2$.
• Calendar Date Product Verification: Choose any calendar month and circle a $2 \times 2$ grid of dates. Compute the product of the diagonals (top-left $\times$ bottom-right, and top-right $\times$ bottom-left). Find the difference between these two products. Repeat with five different months to confirm that the difference is always $7$.