Factoring — common factor extraction, grouping, using identities

2.1.6 · Maths › Algebra — Introduction & Intermediate

Overview

Factoring (ya factorization) ek aisa process hai jisme ek polynomial ko simpler polynomials ke product mein toda jaata hai. Yeh expansion ka ulta hai: instead of $(x + 2) (x + 3) \to x^{2} + 5 x + 6$ , hum ulti direction mein jaate hain: $x^{2} + 5 x + 6 \to (x + 2) (x + 3)$ .

Factor kyun karte hain?

Equations solve karne ke liye: $x^{2} + 5 x + 6 = 0 ⟹ (x + 2) (x + 3) = 0 ⟹ x = - 2$ ya $x = - 3$
Rational expressions simplify karne ke liye: $\frac{x ^{2} - 1}{x - 1} = \frac{( x - 1 ) ( x + 1 )}{x - 1} = x + 1$
Structure reveal karne ke liye: roots dekhna, behavior samajhna, integrate/differentiate karna aasaan ho jaata hai

Core Techniques

1. Common Factor Extraction (Distributive Property in Reverse)

Step-by-step kyun karte hain:

GCF identify karo (Greatest Common Factor) coefficients KA AUR har variable ki lowest power jo sabhi terms mein present ho.
Har term ko GCF se divide karo.
Product ki tarah likhो: GCF times remaining polynomial.

Coefficients: $g cd (6, 9, 15) = 3$
Variables: sabhi terms mein $x$ ki lowest power hai $x^{1}$
GCF = $3 x$

Har term divide karo: $\frac{6 x ^{3}}{3 x} = 2 x^{2}, \frac{9 x ^{2}}{3 x} = 3 x, \frac{- 15 x}{3 x} = - 5$

Result: $6 x^{3} + 9 x^{2} - 15 x = 3 x (2 x^{2} + 3 x - 5)$

Yeh step kyun? $3 x$ bahar nikaalane se remaining polynomial simpler ho jaati hai. Ab zaroorat pade toh $2 x^{2} + 3 x - 5$ ko aur factor kar sakte hain.

Example 2: $4 a^{2} b^{3} - 8 a b^{4} + 12 a^{3} b^{2}$

Coefficients: $g cd (4, 8, 12) = 4$
$a$ : lowest power hai $a^{1}$ (from $- 8 a b^{4}$ )
$b$ : lowest power hai $b^{2}$ (from $12 a^{3} b^{2}$ )
GCF = $4 a b^{2}$

Divide karo: $\frac{4 a ^{2} b ^{3}}{4 a b ^{2}} = ab, \frac{- 8 a b ^{4}}{4 a b ^{2}} = - 2 b^{2}, \frac{12 a ^{3} b ^{2}}{4 a b ^{2}} = 3 a^{2}$

Result: $4 a^{2} b^{3} - 8 a b^{4} + 12 a^{3} b^{2} = 4 a b^{2} (ab - 2 b^{2} + 3 a^{2})$

2. Factoring by Grouping

Kab use karein: Typically 4 terms ke saath, especially rearrange karne ke baad.

Yeh kaam kyun karta hai? Strategically group karke, hum donon groups mein ek common binomial factor $(x + y)$ banate hain, jise phir extract kar sakte hain.

Group karo: $(x^{3} + 3 x^{2}) + (2 x + 6)$

Har group factor karo:

Pehla group: $x^{3} + 3 x^{2} = x^{2} (x + 3)$
Doosra group: $2 x + 6 = 2 (x + 3)$

Ab hamare paas hai: $x^{2} (x + 3) + 2 (x + 3)$

Common factor $(x + 3)$ : $= (x + 3) (x^{2} + 2)$

Yeh step kyun? Har group factor karne ke baad, $(x + 3)$ donon mein dikhta hai. Use extract karne se final factorization milti hai.

Example 4: $2 x y - 3 y + 4 x - 6$

Group karo: $(2 x y - 3 y) + (4 x - 6)$

Factor karo:

$2 x y - 3 y = y (2 x - 3)$
$4 x - 6 = 2 (2 x - 3)$

Result: $y (2 x - 3) + 2 (2 x - 3) = (2 x - 3) (y + 2)$

Example 5 (tricky grouping): $a c + b d + a d + b c$

Standard grouping $(a c + b d) + (a d + b c)$ kaam nahi karti (koi common binomial nahi).

Rearrange karo: $a c + a d + b c + b d$

Group karo: $(a c + a d) + (b c + b d)$

Factor karo:

$a c + a d = a (c + d)$
$b c + b d = b (c + d)$

Result: $(c + d) (a + b)$

Rearrange kyun kiya? Grouping ke liye zaroori hai ki factored pairs ek common binomial share karein. Terms rearrange karne se woh structure ban sakti hai.

3. Using Algebraic Identities

Difference of squares: $a^{2} - b^{2} = (a - b) (a + b)$
Perfect square trinomial (positive): $a^{2} + 2 ab + b^{2} = (a + b)^{2}$
Perfect square trinomial (negative): $a^{2} - 2 ab + b^{2} = (a - b)^{2}$
Sum of cubes: $a^{3} + b^{3} = (a + b) (a^{2} - ab + b^{2})$
Difference of cubes: $a^{3} - b^{3} = (a - b) (a^{2} + ab + b^{2})$

Difference of squares ki derivation:

$(a - b) (a + b)$ se shuru karo:

(a-b)(a+b) &= a \cdot a + a \cdot b - b \cdot a - b \cdot b \\ &= a^2 + ab - ab - b^2 \\ &= a^2 - b^2 \end{align}$$ **Sum of cubes ki derivation:** Hum claim karte hain $a^3 + b^3 = (a+b)(a^2 - ab + b^2)$. RHS expand karo: $$\begin{align} (a+b)(a^2 - ab + b^2) &= a(a^2 - ab + b^2) + b(a^2 - ab + b^2) \\ &= a^3 - a^2b + ab^2 + ba^2 - ab^2 + b^3 \\ &= a^3 + (-a^2b + ba^2) + (ab^2 - ab^2) + b^3 \\ &= a^3 + 0 + b^3 \\ &= a^3 + b^3 \quad \checkmark \end{align}$$ **Yeh kaam kyun karta hai:** Beech ke terms design se cancel ho jaate hain. Cubic identity woh factorization "complete" karti hai jo naive approaches miss kar deti hain. --- > [!example] > **Example 6:** $x^2 - 49$ Pehchano: $x^2 - 7^2$ (difference of squares, $a=x$, $b=7$) $$x^2 - 49 = (x-7)(x+7)$$ --- **Example 7:** $4y^2 - 12y + 9$ Check karo ki perfect square hai ya nahi: $(2y)^2 - 2(2y)(3) + 3^2$? Haan! Yeh $(a-b)^2$ hai jahan $a=2y$, $b=3$. $$4y^2 - 12y + 9 = (2y-3)^2$$ **Check kyun karte hain?** Middle term $-12y = -2(2y)(3)$ pattern confirm karta hai. Agar match na kare, toh alag method use karna padega. --- **Example 8:** $8x^3 + 27$ Pehchano: $8x^3 = (2x)^3$, $27 = 3^3$ → sum of cubes with $a=2x$, $b=3$. $$\begin{align} 8x^3 + 27 &= (2x)^3 + 3^3 \\ &= (2x+3)\left[(2x)^2 - (2x)(3) + 3^2\right] \\ &= (2x+3)(4x^2 - 6x + 9) \end{align}$$ --- **Example 9 (combining techniques):** $x^4 - 16$ Pehle, difference of squares: $(x^2)^2 - 4^2$ $$x^4 - 16 = (x^2 - 4)(x^2 + 4)$$ Lekin $x^2 - 4$ bhi difference of squares hai! $$x^2 - 4 = (x-2)(x+2)$$ **Final factorization:** $$x^4 - 16 = (x-2)(x+2)(x^2+4)$$ (Note: $x^2+4$ real numbers par aur factor nahi hoti, lekin complex numbers par $= (x-2i)(x+2i)$ hoti hai.) **Aur factor kyun karein?** Hamesha check karo ki factors aur factor ho sakte hain ya nahi. Complete factorization saare roots reveal karti hai. --- ![[2.1.06-Factoring-—-common-factor-extraction,-grouping,-using-identities.png]] --- ## Common Mistakes > [!mistake] > **Mistake 1:** Pehle GCF factor out karna bhool jaana. **Galat:** $2x^2 + 8x + 6$ ko directly factor karo... atke! **Steel-man kyun aisa lagta hai:** Aap seedha "product of binomials" patterns par jump karna chahte ho. **Fix:** HAMESHA pehle common factors check karo. $$2x^2 + 8x + 6 = 2(x^2 + 4x + 3) = 2(x+1)(x+3)$$ $2$ miss karne se aage factor karna mushkil ho jaata hai aur kaam adhoora reh jaata hai. --- > [!mistake] > **Mistake 2:** Difference of cubes mein galat signs lagaana. **Galat:** $x^3 - 8 = (x-2)(x^2 - 2x + 4)$ ... nahi ruko, $(x-2)(x^2 + 2x + 4)$? **Steel-man:** Sum vs. difference of cubes ke formulas milte-julte lagte hain aur aasaani se mix ho jaate hain. **Fix:** Sign pattern ke saath memorize karo ya re-derive karo: - **Difference** $a^3 - b^3 = (a-b)(a^2 \mathbf{+} ab + b^2)$ — trinomial mein sign binomial ke sign ka **opposite** hota hai. - **Sum** $a^3 + b^3 = (a+b)(a^2 \mathbf{-} ab + b^2)$ $x^3 - 8= x^3 - 2^3$ ke liye: $$= (x-2)(x^2 + 2x + 4)$$ --- > [!mistake] > **Mistake 3:** Aisi grouping force karna jo kaam nahi karti. **Galat:** $x^3 + 2x^2 + 3x + 5$, group karo $(x^3+2x^2) + (3x+5) = x^2(x+2) + \text{?}(3x+5)$... koi common factor nahi. **Steel-man:** Aap assume karte ho ki saare 4-term polynomials grouping se factor ho jaate hain. **Fix:** Har polynomial nicely factor nahi hoti. Agar alag-alag pairings try karne ke baad bhi grouping fail ho, toh polynomial **prime** (irreducible) ho sakti hai ya advanced techniques ki zaroorat pad sakti hai. Hamesha expand karke apni factorization verify karo. --- > [!mistake] > **Mistake 4:** Galat perfect square identification. **Galat:** $x^2 + 9 = (x+3)^2$? Check karo: $(x+3)^2 = x^2 + 6x + 9 \neq x^2 + 9$. **Steel-man:** Aap do perfect squares dekhte ho aur assume karte ho ki yeh perfect square trinomial hai. **Fix:** Perfect square trinomials mein **teen terms** hote hain: $a^2 \pm 2ab + b^2$. Middle term $2ab$ present aur correct hona chahiye. $x^2 + 9$ actually reals par **prime** hai (complex numbers par $(x-3i)(x+3i)$ ke roop mein factor hoti hai). --- ## Active Recall > [!recall]- > Socho jaise ek 12-saal ke bachche ko explain kar rahe ho: "Factoring ek smoothie un-mix karne jaisi hai. Tumne strawberries aur bananas blend kiye the. Factoring original ingredients dhundti hai. For example, $x^2 + 5x + 6$. Yeh ek 'blended' polynomial hai. Factor karke, hum paate hain ki yeh $(x+2)$ aur $(x+3)$ ko multiply karne se bana tha. Isse kya faayda? Agar tum jaanna chahte ho ki smoothie kab zero ke barabar hogi (solve $x^2+5x+6=0$), toh bas *ek* ingredient zero hona chahiye: $x+2=0$ ya $x+3=0$. Toh $x=-2$ ya $x=-3$. Teen main tricks hain: 1. **Common ingredients nikaalo** (jaise agar har term mein $x$ ho, toh use aage le aao) 2. **Terms ko pairs mein group karo** jo ek pattern share karein 3. **Special recipes pehchano** (jaise $a^2 - b^2$ hamesha $(a-b)(a+b)$ mein split hoti hai) Yeh detective work hai: polynomial dekho, clues dhundho (common factors, patterns), aur reconstruct karo ki yeh kaise bana tha." --- ## Mnemonics & Memory Aids > [!mnemonic] > **SOAP** for difference/sum of cubes: > - **S**ame sign (original problem mein jaisa sign hai, binomial factor ke liye) > - **O**pposite sign (trinomial factor mein, middle term ka) > - **A**lways **P**ositive (trinomial ka last term) **Example:** $a^3 - b^3$ (negative sign) - Binomial: $(a - b)$ — **S**ame (negative) - Trinomial: $a^2 \mathbf{+} ab + b^2$ — **O**pposite (positive), last term **P**ositive **Example:** $a^3 + b^3$ (positive sign) - Binomial: $(a + b)$ — **S**ame (positive) - Trinomial: $a^2 \mathbf{-} ab + b^2$ — **O**pposite (negative), last term **P**ositive --- ## Connections - [[Distributive Property]] — factoring iska reverse hai - [[Polynomial Long Division]] — factorizations verify karna - [[Quadratic Formula]] — jab $ax^2+bx+c$ ke liye factoring fail ho jaaye - [[Rational Root Theorem]] — higher-degree polynomials ke roots dhundna - [[Zero Product Property]] — kyun factoring equations solve karti hai - [[Completing the Square]] — perfect square trinomials pehchaanne ka alternative - [[Complex Numbers]] — $x^2+1$ ya $x^4+4$ jaisi polynomials factor karna - [[Fundamental Theorem of Algebra]] — har polynomial completely factor hoti hai ($\mathbb{C}$ par) --- ## Flashcards #flashcards/maths Algebra mein factoring kya hai? :: Ek polynomial ko simpler polynomials ke product ke roop mein express karne ka process (expansion ka reverse). Kisi bhi factoring technique apply karne se pehle pehla step kya hai? ::: Hamesha greatest common factor (GCF) check karo aur pehle use factor out karo. $a^2 - b^2$ ko factor kaise karte hain? :: $(a-b)(a+b)$ (difference of squares identity). $a^2 + 2ab + b^2$ ki factored form kya hai? ::: $(a+b)^2$ (perfect square trinomial). $a^3 + b^3$ factor karo :: $(a+b)(a^2 - ab + b^2)$ (sum of cubes). $a^3 - b^3$ factor karo ::: $(a-b)(a^2 + ab + b^2)$ (difference of cubes). Factoring by grouping mein, har group factor karne ke baad kya dikhna chahiye? ::: Ek common binomial factor jo donon groups se extract kiya ja sake. $6x^3 - 9x^2 + 12x$ factor karo ::: $3x(2x^2 - 3x + 4)$ — GCF hai $3x$. $x^2 - 25$ factor karo ::: $(x-5)(x+5)$ — difference of squares. $x^2 + 6x + 9$ factor karo ::: $(x+3)^2$ — perfect square trinomial ($a^2+2ab+b^2$ jahan $a=x$, $b=3$). Grouping se factor karo: $xy + 3x + 2y + 6$ ::: $(x+2)(y+3)$ — group karo $(xy+3x)+(2y+6) = x(y+3)+2(y+3)$. $8x^3 - 27$ factor karo ::: $(2x-3)(4x^2+6x+9)$ — difference of cubes jahan $a=2x$, $b=3$. $x^2 + 4$ ko real numbers par prime kyun maana jaata hai? ::: Yeh sum of squares hai bina middle term ke; yeh kisi real factoring pattern mein fit nahi hota (complex par factors: $(x-2i)(x+2i)$). Cubes ke liye SOAP mnemonic kya hai? ::: Binomial mein **S**ame sign, trinomial ke middle term mein **O**pposite, last term **A**lways **P**ositive. Completely factor karo: $2x^3 + 8x^2 + 8x$ ::: $2x(x^2+4x+4) = 2x(x+2)^2$ — pehle GCF factor karo, phir perfect square pehchano. ## 🖼️ Concept Map ```mermaid flowchart TD F[Factoring polynomials] E[Expansion] T1[Common factor extraction] T2[Factoring by grouping] T3[Using identities] GCF[Find GCF] GRP[Group terms in pairs] BIN[Common binomial factor] SOLVE[Solve equations] SIMP[Simplify rationals] F -->|reverse of| E F -->|technique| T1 F -->|technique| T2 F -->|technique| T3 T1 -->|requires| GCF T2 -->|first step| GRP GRP -->|reveals| BIN BIN -->|extracted gives| T2 F -->|used to| SOLVE F -->|used to| SIMP SOLVE -->|via| BIN ```