🕉️ VEDIC MATHEMATICS — LEVEL 2: INTERMEDIATE

MODULE 16: Factorization & Algebraic Products

Complete Study Material | Theory + Examples + Practice + Test Bank

"Factorization is the art of seeing the whole as a product of its parts. The sutras reveal the hidden structure within algebraic expressions." — Vedic Mathematics Teacher's Manual

📋 MODULE AT A GLANCE

🎯 LEARNING OUTCOMES

By the end of this module, the student will be able to:

1. Factor quadratic expressions in under 10 seconds using the Vedic middle-term split method
2. Apply the Adyamadyena (first by first, last by last) method for factoring special cases
3. Multiply any two binomials using Urdhva-Tiryagbhyam mentally
4. Multiply two trinomials using the extended Urdhva pattern
5. Factor cubic polynomials using the Paravartya (transpose) method
6. Find the HCF of polynomials using the Vedic method (Chalana-Kalanabhyam)
7. Verify factorization results using Gunitasamuccayah (sum of coefficients)
8. Recognize the relationship between coefficients and factors using Sutra 16

PART 1: THEORY

1.1 — Introduction to Vedic Factorization

What is Factorization?

Factorization is the process of breaking down an algebraic expression into a product of simpler expressions (factors).

Example: $x^2 + 5x + 6 = (x + 2)(x + 3)$

Why Vedic Factorization is Different

1.2 — Sutras Used in This Module

1.3 — Factoring Quadratics: The Vedic Method

Standard Form

For a quadratic expression: $ax^2 + bx + c$ (where a ≠ 0)

The Vedic Principle

Find two numbers whose:

• Product = $a \times c$
• Sum = $b$

Then split the middle term $bx$ into the sum of these two numbers times x, and factor by grouping.

Example 1: $x^2 + 7x + 12$

Here a=1, b=7, c=12

Product = a×c = 1×12 = 12 Sum = b = 7

Find two numbers with product 12 and sum 7: 3 and 4

Therefore: $x^2 + 7x + 12 = (x + 3)(x + 4)$ ✓

Example 2: $x^2 + 5x + 6$

Product = 6, Sum = 5 → Numbers: 2 and 3

$x^2 + 5x + 6 = (x + 2)(x + 3)$ ✓

Example 3: $x^2 - 5x + 6$

Product = 6, Sum = -5 → Numbers: -2 and -3 (since -2 × -3 = 6, -2 + -3 = -5)

$x^2 - 5x + 6 = (x - 2)(x - 3)$ ✓

Example 4: $x^2 + x - 6$

Product = -6, Sum = 1 → Numbers: 3 and -2 (3 × -2 = -6, 3 + (-2) = 1)

$x^2 + x - 6 = (x + 3)(x - 2)$ ✓

Example 5: $2x^2 + 7x + 3$

Here a=2, b=7, c=3

Product = a × c = 2 × 3 = 6 Sum = b = 7

Find two numbers with product 6 and sum 7: 1 and 6

Split middle term: $2x^2 + 1x + 6x + 3$ Group: $x(2x + 1) + 3(2x + 1) = (2x + 1)(x + 3)$ ✓

Example 6: $6x^2 + 13x + 6$

Product = 6 × 6 = 36 Sum = 13 Numbers: 4 and 9 (4×9=36, 4+9=13)

Split: $6x^2 + 4x + 9x + 6$ Group: $2x(3x + 2) + 3(3x + 2) = (3x + 2)(2x + 3)$ ✓

Example 7: $4x^2 - 19x + 12$

Product = 4 × 12 = 48 Sum = -19 Numbers: -3 and -16 (-3 × -16 = 48, -3 + -16 = -19)

Split: $4x^2 - 3x - 16x + 12$ Group: $x(4x - 3) - 4(4x - 3) = (4x - 3)(x - 4)$ ✓

1.4 — Adyamadyena Method (First by First, Last by Last)

Sub-Sutra: Adyamadyena Antyamantyena

Application to Factoring

For a quadratic $ax^2 + bx + c$, if it factors into $(px + q)(rx + s)$, then:

• $p \times r = a$ (first by first)
• $q \times s = c$ (last by last)
• $p \times s + q \times r = b$ (cross terms)

This gives us a systematic search method:

Step 1: Find factor pairs of a (p and r) Step 2: Find factor pairs of c (q and s) Step 3: Check which combination gives $ps + qr = b$

Example: $6x^2 + 13x + 6$

Factor pairs of 6 (a): (1,6), (2,3), (3,2), (6,1) Factor pairs of 6 (c): (1,6), (2,3), (3,2), (6,1)

Test (p=2, r=3) and (q=3, s=2): Cross terms: $p \times s + q \times r = 2×2 + 3×3 = 4 + 9 = 13$ ✓

Thus: $(2x + 3)(3x + 2)$ ✓

1.5 — Product of Two Binomials Using Urdhva

The Urdhva Pattern for Binomials

For $(px + q)(rx + s)$:

Example 1: $(2x + 3)(4x + 5)$

• $x^2$: $2×4 = 8$
• $x$: $2×5 + 3×4 = 10 + 12 = 22$
• Constant: $3×5 = 15$

Result: $8x^2 + 22x + 15$

Example 2: $(3x - 2)(5x + 4)$

• $x^2$: $3×5 = 15$
• $x$: $3×4 + (-2)×5 = 12 - 10 = 2$
• Constant: $(-2)×4 = -8$

Result: $15x^2 + 2x - 8$

Example 3: $(x + 3)(x - 5)$

• $x^2$: $1×1 = 1$
• $x$: $1×(-5) + 3×1 = -5 + 3 = -2$
• Constant: $3×(-5) = -15$

Result: $x^2 - 2x - 15$

1.6 — Product of Two Trinomials Using Urdhva

The 3×3 Urdhva Pattern

For $(a_2x^2 + a_1x + a_0)(b_2x^2 + b_1x + b_0)$:

Example: $(x^2 + 2x + 3)(x^2 + 4x + 5)$

Result: $x^4 + 6x^3 + 16x^2 + 22x + 15$

Example 2: $(2x^2 + 3x + 4)(x^2 + 5x + 6)$

Result: $2x^4 + 13x^3 + 31x^2 + 38x + 24$

1.7 — Factoring Cubics: The Paravartya Method

Sutra 4: Paravartya Yojayet (Transpose and Apply)

For factoring a cubic expression $ax^3 + bx^2 + cx + d$, we can use the factor theorem:

Step 1: Find a factor using the rational root theorem (possible roots = factors of d / factors of a)

Step 2: Use synthetic division (Paravartya) to divide the cubic by the factor

Step 3: Factor the resulting quadratic

Example 1: $x^3 - 6x^2 + 11x - 6$

Possible roots (factors of 6): ±1, ±2, ±3, ±6

Test x=1: $1 - 6 + 11 - 6 = 0$ ✓ (x-1) is a factor

Use Paravartya (synthetic division):

Coefficients: 1, -6, 11, -6 Bring down 1 1 × 1 = 1 → add to -6 = -5 -5 × 1 = -5 → add to 11 = 6 6 × 1 = 6 → add to -6 = 0

Quotient: $x^2 - 5x + 6 = (x - 2)(x - 3)$

Therefore: $x^3 - 6x^2 + 11x - 6 = (x - 1)(x - 2)(x - 3)$ ✓

Example 2: $x^3 - 3x^2 - 4x + 12$

Possible roots (factors of 12): ±1, ±2, ±3, ±4, ±6, ±12

Test x=2: $8 - 12 - 8 + 12 = 0$ ✓ (x-2) is a factor

Synthetic division with root 2:

Coefficients: 1, -3, -4, 12 Bring down 1 1×2=2 → add to -3 = -1 -1×2=-2 → add to -4 = -6 -6×2=-12 → add to 12 = 0

Quotient: $x^2 - x - 6 = (x - 3)(x + 2)$

Therefore: $x^3 - 3x^2 - 4x + 12 = (x - 2)(x - 3)(x + 2)$ ✓

Example 3: $2x^3 + 3x^2 - 23x - 12$

Possible roots (factors of -12 / factors of 2): ±1, ±2, ±3, ±4, ±6, ±12, ±1/2, ±3/2, etc.

Test x=3: $2×27 + 3×9 - 23×3 - 12 = 54 + 27 - 69 - 12 = 0$ ✓ (x-3) is a factor

Synthetic division with root 3:

Coefficients: 2, 3, -23, -12 Bring down 2 2×3=6 → add to 3 = 9 9×3=27 → add to -23 = 4 4×3=12 → add to -12 = 0

Quotient: $2x^2 + 9x + 4 = (2x + 1)(x + 4)$

Therefore: $2x^3 + 3x^2 - 23x - 12 = (x - 3)(2x + 1)(x + 4)$ ✓

1.8 — HCF of Polynomials Using Vedic Method

Sutra 9: Chalana-Kalanabhyam (Differences)

For finding the HCF (Highest Common Factor) of two polynomials:

Method: Successively subtract multiples of one polynomial from the other until the remainder is a factor of both.

Example: Find HCF of $x^2 + 5x + 6$ and $x^2 + 4x + 3$

Subtract: $(x^2 + 5x + 6) - (x^2 + 4x + 3) = x + 3$

Check if (x+3) divides both:

• $x^2 + 5x + 6 = (x+3)(x+2)$ ✓
• $x^2 + 4x + 3 = (x+3)(x+1)$ ✓

HCF = $x + 3$

Example: Find HCF of $x^3 - x^2 - 5x - 3$ and $x^2 - 4x - 5$

First, factor each if possible. Or use repeated subtraction.

Better method: Use the Vedic method of successive differences.

But for polynomials, factoring is often simpler. Let me demonstrate the systematic method:

Step 1: Divide the larger-degree polynomial by the smaller-degree polynomial.

Divide $x^3 - x^2 - 5x - 3$ by $x^2 - 4x - 5$:

Quotient ≈ x, remainder = $(x^3 - x^2 - 5x - 3) - x(x^2 - 4x - 5) = x^3 - x^2 - 5x - 3 - x^3 + 4x^2 + 5x = 3x^2 + 0x - 3 = 3(x^2 - 1)$

Now find HCF of $x^2 - 4x - 5$ and $x^2 - 1$

Subtract: $(x^2 - 4x - 5) - (x^2 - 1) = -4x - 4 = -4(x + 1)$

Check if (x+1) divides both:

• $x^2 - 1 = (x+1)(x-1)$ ✓
• $x^2 - 4x - 5 = (x+1)(x-5)$ ✓

HCF = $x + 1$

1.9 — Gunitasamuccayah: Verification of Factorization

Sutra 15: Gunitasamuccayah

"The product of the sum equals the sum of the products"

For factorization verification:

• Sum of coefficients of the original polynomial = Product of (sum of coefficients of each factor)

Example 1: Verify $(x + 3)(x + 4) = x^2 + 7x + 12$

Sum of coefficients of original: $1 + 7 + 12 = 20$ Sum of coefficients of (x+3): $1 + 3 = 4$ Sum of coefficients of (x+4): $1 + 4 = 5$ Product: $4 × 5 = 20$ ✓

Example 2: Verify $(2x + 1)(x + 3) = 2x^2 + 7x + 3$

Original sum: $2 + 7 + 3 = 12$ Factor sums: $(2+1)=3$, $(1+3)=4$ Product: $3 × 4 = 12$ ✓

Example 3: Verify $(x-2)(x+3) = x^2 + x - 6$

Original sum: $1 + 1 + (-6) = -4$ Factor sums: $(1-2)=-1$, $(1+3)=4$ Product: $(-1) × 4 = -4$ ✓

1.10 — Gunakasamuccayah: The Companion Sutra

Sutra 16: Gunakasamuccayah

"The factors of the sum equals the sum of the factors"

This is the converse of Sutra 15. It states that the factors of the sum of coefficients correspond to the sums of coefficients of the factors.

Application

For a polynomial $P(x)$ with factors $F_1(x), F_2(x), ..., F_n(x)$:

$$\text{Sum of coefficients of } P = \prod_{i=1}^n (\text{Sum of coefficients of } F_i)$$

This provides a powerful verification tool, especially when factoring polynomials with more than two factors.

Example: $x^3 - 6x^2 + 11x - 6 = (x-1)(x-2)(x-3)$

Original sum: $1 - 6 + 11 - 6 = 0$ Factor sums: $(1-1)=0$, $(1-2)=-1$, $(1-3)=-2$ Product: $0 × (-1) × (-2) = 0$ ✓

PART 2: WORKED EXAMPLES

Section A: Factoring Quadratics

Example 1

Question: Factor $x^2 + 9x + 20$.

Answer:

Product = 20, Sum = 9 → Numbers: 4 and 5

$x^2 + 9x + 20 = (x + 4)(x + 5)$ ✓

Example 2

Question: Factor $x^2 - 8x + 15$.

Answer:

Product = 15, Sum = -8 → Numbers: -3 and -5

$x^2 - 8x + 15 = (x - 3)(x - 5)$ ✓

Example 3

Question: Factor $x^2 - 2x - 15$.

Answer:

Product = -15, Sum = -2 → Numbers: -5 and 3

$x^2 - 2x - 15 = (x - 5)(x + 3)$ ✓

Example 4

Question: Factor $3x^2 + 10x + 8$.

Answer:

Product = 3 × 8 = 24, Sum = 10 → Numbers: 4 and 6

Split: $3x^2 + 4x + 6x + 8$ Group: $x(3x + 4) + 2(3x + 4) = (3x + 4)(x + 2)$ ✓

Example 5

Question: Factor $6x^2 - 5x - 6$.

Answer:

Product = 6 × (-6) = -36, Sum = -5 → Numbers: -9 and 4

Split: $6x^2 - 9x + 4x - 6$ Group: $3x(2x - 3) + 2(2x - 3) = (2x - 3)(3x + 2)$ ✓

Example 6

Question: Factor $4x^2 + 12x + 9$.

Answer:

Product = 4 × 9 = 36, Sum = 12 → Numbers: 6 and 6 (perfect square)

Split: $4x^2 + 6x + 6x + 9$ Group: $2x(2x + 3) + 3(2x + 3) = (2x + 3)^2$ ✓

Example 7

Question: Factor $8x^2 + 2x - 15$.

Answer:

Product = 8 × (-15) = -120, Sum = 2 → Numbers: 12 and -10

Split: $8x^2 + 12x - 10x - 15$ Group: $4x(2x + 3) - 5(2x + 3) = (2x + 3)(4x - 5)$ ✓

Section B: Product of Binomials

Example 8

Question: Multiply $(3x + 5)(2x + 7)$ using Urdhva.

Answer:

• $x^2$: $3×2 = 6$
• $x$: $3×7 + 5×2 = 21 + 10 = 31$
• Constant: $5×7 = 35$

Result: $6x^2 + 31x + 35$

Example 9

Question: Multiply $(4x - 3)(2x + 5)$.

Answer:

• $x^2$: $4×2 = 8$
• $x$: $4×5 + (-3)×2 = 20 - 6 = 14$
• Constant: $(-3)×5 = -15$

Result: $8x^2 + 14x - 15$

Example 10

Question: Multiply $(x + 2)(x^2 + 3x + 4)$.

Answer: (Binomial × Trinomial)

Use distributive or Urdhva with padding: treat binomial as (0x² + 1x + 2)

Result: $x^3 + 5x^2 + 10x + 8$

Section C: Product of Trinomials

Example 11

Question: Multiply $(2x^2 + x + 3)(x^2 + 2x + 4)$.

Answer:

Result: $2x^4 + 5x^3 + 13x^2 + 10x + 12$

Example 12

Question: Multiply $(x^2 - x + 2)(x^2 + x + 3)$.

Answer:

Result: $x^4 + 4x^2 - x + 6$

Section D: Factoring Cubics

Example 13

Question: Factor $x^3 - 3x^2 - x + 3$.

Answer:

Possible roots: ±1, ±3

Test x=1: $1 - 3 - 1 + 3 = 0$ ✓ (x-1) is factor

Synthetic division with root 1:

Coefficients: 1, -3, -1, 3 Bring down 1 1×1=1 → add to -3 = -2 -2×1=-2 → add to -1 = -3 -3×1=-3 → add to 3 = 0

Quotient: $x^2 - 2x - 3 = (x - 3)(x + 1)$

Therefore: $x^3 - 3x^2 - x + 3 = (x - 1)(x - 3)(x + 1)$

Example 14

Question: Factor $2x^3 + 5x^2 - 4x - 3$.

Answer:

Possible roots: ±1, ±3, ±1/2, ±3/2

Test x=1: $2 + 5 - 4 - 3 = 0$ ✓ (x-1) is factor

Synthetic division:

Coefficients: 2, 5, -4, -3 Bring down 2 2×1=2 → add to 5 = 7 7×1=7 → add to -4 = 3 3×1=3 → add to -3 = 0

Quotient: $2x^2 + 7x + 3 = (2x + 1)(x + 3)$

Therefore: $2x^3 + 5x^2 - 4x - 3 = (x - 1)(2x + 1)(x + 3)$

Example 15

Question: Factor $x^3 - 2x^2 - 5x + 6$.

Answer:

Possible roots: ±1, ±2, ±3, ±6

Test x=1: $1 - 2 - 5 + 6 = 0$ ✓

Synthetic division with root 1:

Coefficients: 1, -2, -5, 6 Bring down 1 1×1=1 → add to -2 = -1 -1×1=-1 → add to -5 = -6 -6×1=-6 → add to 6 = 0

Quotient: $x^2 - x - 6 = (x - 3)(x + 2)$

Therefore: $x^3 - 2x^2 - 5x + 6 = (x - 1)(x - 3)(x + 2)$

Section E: Verification Using Gunitasamuccayah

Example 16

Question: Verify that $(3x + 1)(x - 2) = 3x^2 - 5x - 2$.

Answer:

Original sum: $3 + (-5) + (-2) = -4$ Factor sums: $(3+1)=4$, $(1-2)=-1$ Product: $4 × (-1) = -4$ ✓

Example 17

Question: Verify that $(2x - 3)(x + 4)(x - 1) = 2x^3 + 3x^2 - 17x + 12$.

Answer:

Original sum: $2 + 3 + (-17) + 12 = 0$ Factor sums: $(2-3)=-1$, $(1+4)=5$, $(1-1)=0$ Product: $(-1) × 5 × 0 = 0$ ✓

PART 3: PRACTICE EXERCISES

Exercise Set A: Factoring Quadratics (20 Questions)

Factor each quadratic expression.

A1. $x^2 + 8x + 12$ A2. $x^2 + 10x + 21$ A3. $x^2 + 6x + 8$ A4. $x^2 - 7x + 12$ A5. $x^2 - 9x + 20$ A6. $x^2 - 5x + 6$ A7. $x^2 + 2x - 15$ A8. $x^2 - 3x - 10$ A9. $x^2 - 4x - 12$ A10. $x^2 + 5x - 24$ A11. $2x^2 + 5x + 2$ A12. $3x^2 + 10x + 3$ A13. $4x^2 + 11x + 6$ A14. $5x^2 + 14x + 8$ A15. $2x^2 - 5x - 3$ A16. $3x^2 - 8x - 3$ A17. $4x^2 - 11x - 3$ A18. $6x^2 - 7x - 5$ A19. $9x^2 + 12x + 4$ A20. $16x^2 - 24x + 9$

Exercise Set B: Product of Binomials (15 Questions)

Multiply using Urdhva.

B1. $(x + 4)(x + 7)$ B2. $(x - 3)(x + 5)$ B3. $(2x + 3)(3x + 4)$ B4. $(3x - 2)(2x + 5)$ B5. $(4x + 1)(x - 3)$ B6. $(5x - 2)(3x - 4)$ B7. $(x + 2)(x^2 + 3x + 4)$ B8. $(2x + 1)(x^2 + 4x + 3)$ B9. $(x - 3)(2x^2 + x + 5)$ B10. $(3x - 1)(x^2 - 2x + 4)$ B11. $(x^2 + 2x + 1)(x + 1)$ B12. $(x^2 - x + 1)(x + 2)$ B13. $(2x^2 + 3x)(x - 1)$ B14. $(3x^2 - 2x + 1)(2x - 1)$ B15. $(x^2 + x + 1)(x^2 - x + 1)$

Exercise Set C: Product of Trinomials (10 Questions)

Multiply using the 3×3 Urdhva pattern.

C1. $(x^2 + 3x + 2)(x^2 + 4x + 3)$ C2. $(x^2 + x + 1)(x^2 + 2x + 3)$ C3. $(2x^2 + x + 3)(x^2 + 2x + 1)$ C4. $(3x^2 + 2x + 1)(x^2 + x + 2)$ C5. $(x^2 - 2x + 1)(x^2 + 2x + 1)$ C6. $(2x^2 + 3x + 4)(x^2 + 2x + 3)$ C7. $(x^2 + 2x + 3)(x^2 + 2x + 3)$ (square) C8. $(2x^2 - x + 3)(x^2 + 4x - 1)$ C9. $(x^2 + 5x + 6)(x^2 + 5x + 6)$ C10. $(3x^2 - 2x - 1)(2x^2 + x - 3)$

Exercise Set D: Factoring Cubics (10 Questions)

Factor each cubic completely.

D1. $x^3 - 4x^2 + x + 6$ D2. $x^3 - 6x^2 + 11x - 6$ D3. $x^3 - 3x^2 - 4x + 12$ D4. $x^3 + 2x^2 - 5x - 6$ D5. $x^3 - x^2 - 8x + 12$ D6. $2x^3 + 3x^2 - 11x - 6$ D7. $3x^3 - 5x^2 - 4x + 4$ D8. $4x^3 + 4x^2 - 11x - 6$ D9. $x^3 - 7x - 6$ D10. $x^3 + 4x^2 + x - 6$

Exercise Set E: HCF of Polynomials (5 Questions)

Find the HCF of each pair of polynomials.

E1. $x^2 + 5x + 4$ and $x^2 + 3x + 2$ E2. $x^2 - 4x + 3$ and $x^2 - 5x + 4$ E3. $x^3 - 1$ and $x^2 + x + 1$ E4. $x^3 + 2x^2 - x - 2$ and $x^2 + 3x + 2$ E5. $x^4 - 1$ and $x^2 - 1$

Exercise Set F: Verification (10 Questions)

Verify each factorization using Gunitasamuccayah (sum of coefficients).

F1. $(x + 2)(x + 5) = x^2 + 7x + 10$ F2. $(x - 3)(x + 4) = x^2 + x - 12$ F3. $(2x + 1)(x + 3) = 2x^2 + 7x + 3$ F4. $(3x - 2)(x - 4) = 3x^2 - 14x + 8$ F5. $(x + 1)(x - 2)(x + 3) = x^3 + 2x^2 - 5x - 6$ F6. $(x - 1)(x - 2)(x - 3) = x^3 - 6x^2 + 11x - 6$ F7. $(2x - 1)(x + 2)(x - 3) = 2x^3 - 3x^2 - 11x + 6$ F8. $(x + 1)^3 = x^3 + 3x^2 + 3x + 1$ F9. $(x^2 + x + 1)(x + 1) = x^3 + 2x^2 + 2x + 1$ F10. $(2x^2 + 3x + 4)(x^2 + 5x + 6) = 2x^4 + 13x^3 + 31x^2 + 38x + 24$

Answer Key for Practice Exercises

Set A Answers:

A1. (x+2)(x+6)
A2. (x+3)(x+7)
A3. (x+2)(x+4)
A4. (x-3)(x-4)
A5. (x-4)(x-5)
A6. (x-2)(x-3)
A7. (x+5)(x-3)
A8. (x-5)(x+2)
A9. (x-6)(x+2)
A10. (x+8)(x-3)
A11. (2x+1)(x+2)
A12. (3x+1)(x+3)
A13. (4x+3)(x+2)
A14. (5x+4)(x+2)
A15. (2x+1)(x-3)
A16. (3x+1)(x-3)
A17. (4x+1)(x-3)
A18. (3x-5)(2x+1)
A19. (3x+2)²
A20. (4x-3)²

Set B Answers:

B1. $x^2 + 11x + 28$
B2. $x^2 + 2x - 15$
B3. $6x^2 + 17x + 12$
B4. $6x^2 + 11x - 10$
B5. $4x^2 - 11x - 3$
B6. $15x^2 - 26x + 8$
B7. $x^3 + 5x^2 + 10x + 8$
B8. $2x^3 + 9x^2 + 10x + 3$
B9. $2x^3 - 5x^2 + 2x - 15$
B10. $3x^3 - 7x^2 + 14x - 4$
B11. $x^3 + 3x^2 + 3x + 1$
B12. $x^3 + x^2 - x + 2$
B13. $2x^3 + x^2 - 3x$
B14. $6x^3 - 7x^2 + 4x - 1$
B15. $x^4 + x^2 + 1$

Set C Answers:

C1. $x^4 + 7x^3 + 18x^2 + 17x + 6$
C2. $x^4 + 3x^3 + 6x^2 + 5x + 3$
C3. $2x^4 + 5x^3 + 7x^2 + 7x + 3$
C4. $3x^4 + 5x^3 + 9x^2 + 5x + 2$
C5. $x^4 - 2x^2 + 1$
C6. $2x^4 + 7x^3 + 16x^2 + 17x + 12$
C7. $x^4 + 4x^3 + 10x^2 + 12x + 9$
C8. $2x^4 + 7x^3 - 14x^2 + 13x - 3$
C9. $x^4 + 10x^3 + 37x^2 + 60x + 36$
C10. $6x^4 - x^3 - 14x^2 + 3x + 3$

Set D Answers:

D1. (x-1)(x-2)(x+3)
D2. (x-1)(x-2)(x-3)
D3. (x-2)(x-3)(x+2)
D4. (x+1)(x-2)(x+3)
D5. (x-2)(x-2)(x+3) = (x-2)²(x+3)
D6. (x+2)(2x+1)(x-3)
D7. (x-1)(3x+2)(x-2)
D8. (x+2)(4x+3)(x-1)
D9. (x+1)(x+2)(x-3)
D10. (x+3)(x+1)(x-2)

Set E Answers:

E1. x+1
E2. x-1
E3. x²+x+1
E4. x+2
E5. x²-1

Set F Answers:

All verifications pass (sum of coefficients match).

PART 5: TEACHER'S GUIDE

Common Mistakes & Corrections

Memory Aid for Factor Theorem

QUICK REFERENCE CARD

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║              MODULE 16 — FACTORIZATION & ALGEBRAIC PRODUCTS            ║
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║  FACTORING QUADRATICS (ax² + bx + c):                                 ║
║  ┌─────────────────────────────────────────────────────────────┐      ║
║  │ Find two numbers with product = a×c and sum = b             │      ║
║  │ Split middle term → factor by grouping                      │      ║
║  └─────────────────────────────────────────────────────────────┘      ║
║                                                                       ║
║  URDHAVA FOR BINOMIALS: (px+q)(rx+s) = pr x² + (ps+qr)x + qs          ║
║                                                                       ║
║  URDHAVA FOR TRINOMIALS: 3×3 pattern (5 steps: 1,2,3,2,1 products)    ║
║                                                                       ║
║  PARAVARTYA FOR CUBICS: Synthetic division with possible roots        ║
║                                                                       ║
║  GUNITASAMUCCAYAH (Verification):                                     ║
║  Sum of coefficients of product = Product of sums of coefficients     ║
║                                                                       ║
║  SUTRA 3: Urdhva-Tiryagbhyam — Vertically and cross-wise              ║
║  SUTRA 4: Paravartya Yojayet — Transpose and apply                    ║
║  SUTRA 15: Gunitasamuccayah — Product of sums = sum of products       ║
║  SUTRA 16: Gunakasamuccayah — Factors of sum = sum of factors         ║
║                                                                       ║
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Document Version 1.0 | Vedic Mathematics Level 2 Intermediate Course

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