1. Factorisation Kya Hai? 🤔
Factorisation ka matlab hai kisi badi cheez ko uske chhote parts (factors) mein todna, jinko multiply karne par wapas wahi badi cheez mil jaye.
Imagine karo aapne ek bada Lego Castle banaya hai (Polynomial). Ab aapko dekhna hai ki ye bana kaise hai.
Aap usse tod kar chhote-chhote blocks (Factors) mein alag karte ho. Jab aap in blocks ko wapas jodoge (Multiply karoge), toh wahi Castle wapas ban jayega.
Bas yahi Factorisation hai - Polynomial ko uske basic building blocks mein todna!
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📊 Infographic Summary
2. Factor Theorem 📜
Factor Theorem ek simple rule hai jo batata hai ki koi number polynomial ka factor hai ya nahi.
Theorem: Agar p(a) = 0 hai, toh (x - a) polynomial
p(x) ka ek factor hai.
Ulta bhi sahi hai: Agar (x - a) factor hai, toh p(a) = 0 hoga.
Socho har lock (Polynomial) ki ek specific key (Factor) hoti hai.
Agar aap key lagate ho aur lock khul jata hai (Remainder 0 aata hai), toh wo key sahi hai!
Yahan x = a wo key hai. Agar value put karne par answer 0 aaye, toh (x - a)
sahi key (factor) hai.
Example: Check if x - 2 is a factor of p(x) = x³ - 8
Solution:
Yahan x - 2 check karna hai, toh hum x = 2 put karenge.
p(2) = (2)³ - 8
= 8 - 8 = 0
Since answer 0 hai, Yes, (x - 2) is a factor! ✅
3. Factorising Cubic Polynomials
Cubic polynomials (degree 3) ko factorise karne ke liye humein thoda smart work karna padta hai. Hum pehle ek factor guess karte hain, aur phir baaki ke factors nikalte hain.
Example 10 : Factorise x³ – 23x² + 142x – 120.
Solution:
Let p(x) = x³ – 23x² + 142x – 120
Sabse pehle, hum constant term -120 ke factors dekhte hain. Ye ho sakte hain: ±1, ±2, ±3, ±4, ±5, ±6, ±8, ±10, ±12, ±15, ±20, ±24, ±30, ±60.
Step 1: Trial Method (Hit and Trial)
Hum x = 1 try karte hain:
p(1) = (1)³ – 23(1)² + 142(1) – 120
= 1 – 23 + 142 – 120
= 143 – 143 = 0
Kyunki p(1) = 0, iska matlab (x – 1) ek factor hai. (Factor
Theorem)
Step 2: Division or Adjustment
Ab hum polynomial ko adjust karke likhte hain taaki (x - 1) common aa jaye:
x³ – 23x² + 142x – 120
= x³ – x² – 22x² + 22x + 120x – 120
= x²(x – 1) – 22x(x – 1) + 120(x – 1)
= (x – 1) (x² – 22x + 120) [Taking (x – 1) common]
Alternatively, aap p(x) ko (x - 1) se long division karke bhi x² – 22x + 120
nikaal
sakte hain.
Step 3: Factorising the Quadratic Part
Ab humein x² – 22x + 120 ko factorise karna hai. Isse hum "Splitting the Middle
Term" se
kar sakte hain.
Humein aise do numbers chahiye jinka sum -22 ho aur product 120 ho. Wo numbers hain -12 aur -10.
x² – 22x + 120
= x² – 12x – 10x + 120
= x(x – 12) – 10(x – 12)
= (x – 12) (x – 10)
Final Answer:
So, x³ – 23x² – 142x – 120 = (x – 1)(x – 10)(x – 12)
🚀 Next Steps
Factorisation master kar liya? Awesome! Ab next lesson mein hum seekhenge Algebraic Identities jo calculations ko super fast bana deti hain.