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व्यष्टिसमष्टि

Vyaṣṭisamaṣṭhi

"Part and whole"

What this sutra solves

Factor a quadratic quickly — the basis for solving, simplifying, and finding dimensions.

A rectangular garden has area x² + 5x + 6. You want its two side lengths.

becomesFactor x2+5x+6x^2 + 5x + 6

Two numbers with product 6 and sum 5 are 2 and 3 → sides (x+2) and (x+3).

Live Demo
Factor the trinomial

Vyaṣṭisamaṣṭhi balances the "individual" (the parts) with the "whole." For x² + bx + c, we need two parts that fit both the product and the sum.

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Factor the trinomial

Vyaṣṭisamaṣṭhi balances the "individual" (the parts) with the "whole." For x² + bx + c, we need two parts that fit both the product and the sum.

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⚡ Speed Advantage

Vedic
5 steps
Traditional
6 steps

See how few steps Vedic method needs!

Best for

  • Factoring quadratic and higher-degree polynomials

Use when

  • Quadratic trinomials, factorization problems

Avoid when

  • Irreducible polynomials over integers

Intuition

Factor a quadratic by finding two numbers whose sum is the middle coefficient and product is the last — the Vedic way to factor trinomials.

Story Mode

From Whole to Parts

Every trinomial is a product hiding its factors. The 'whole' (samasthi) is the expression; the 'parts' (vyashti) are the two root-factors. Sum and product are the two clues. Find what multiplies to the constant and adds to the middle — the factors reveal themselves.

Vedic vs conventional

Conventional

trial and error factoring — variable steps.

Vedic

systematic sum-and-product recognition.

Structured search replaces random trial-and-error.

Applications

Factoring quadratic trinomials

Factor x²+bx+c by finding two numbers summing to b with product c.

x2+5x+6x^2+5x+6x2+4x21x^2+4x-216x2+5x66x^2+5x-6

Common Mistakes to Avoid

Using product but not sum to choose factors

Wrong approach

For x2+4x21x^2+4x-21, choosing 3 and −7 because their product is −21, even though their sum is −4.

Correct approach

The two numbers must satisfy both conditions: product =21= -21 and sum =4= 4, so use 7 and −3.

Why this happens

💡 Students latch onto the constant term and forget the middle coefficient.

Dropping the leading coefficient in non-monic quadratics

Wrong approach

Factoring 6x2+5x66x^2+5x-6 as if it were x2+5x6x^2+5x-6.

Correct approach

When the leading coefficient is not 1, account for it while splitting the middle term or grouping.

Why this happens

💡 The beginner pattern x²+bx+c is over-applied to harder quadratics.

Why It Works

Start from two linear factors:

(x+a)(x+b)(x+a)(x+b)

Expand:

=x2+(a+b)x+ab=x^2+(a+b)x+ab

Read the two clues:

sum=a+b,product=ab\text{sum}=a+b,\quad \text{product}=ab

Factoring reverses expansion: find two parts whose sum gives the middle coefficient and whose product gives the constant term.

x²+(a+ba+b)x+ab = (x+ax+a)(x+bx+b). The sutra recognizes that the 'individual' factors (vyashti) combine into the 'whole' (samasthi) through their sum and product.

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