Remainder theorem and factor theorem — proof and applications

2.1.15 · Maths › Algebra — Introduction & Intermediate

The Remainder Theorem

Derivation from First Principles

Chaliye yeh step-by-step prove karte hain, kyun har step kaam karta hai yeh samajhte hue.

Starting Point: Jab hum kisi bhi polynomial $p (x)$ ko $(x - a)$ se divide karte hain, hum likh sakte hain:

$p (x) = (x - a) \cdot q (x) + r$

Jahaan:

$q (x)$ quotient polynomial hai
$r$ remainder hai (jo ek constant hona chahiye kyunki hum ek linear term se divide kar rahe hain)

Remainder constant kyun hota hai? Remainder ki degree divisor ki degree se kam honi chahiye. Kyunki $(x - a)$ ki degree 1 hai, remainder ki degree 0 (ek constant) honi chahiye ya phir woh zero hoga.

The Key Insight: Yeh equation x ki saari values ke liye sach honi chahiye. Toh chaliye $x = a$ substitute karte hain:

$p (a) = (a - a) \cdot q (a) + r$

Yeh step kyun? Humne specifically $x = a$ isliye choose kiya taaki $(x - a)$ wala term gayab ho jaaye!

$p (a) = 0 \cdot q (a) + r$

$p (a) = r$

Conclusion: $p (x)$ ko $(x - a)$ se divide karne par remainder exactly $p (a)$ hota hai. ∎

Figure — Remainder theorem and factor theorem — proof and applications

The Factor Theorem

Derivation from Remainder Theorem

Factor Theorem, Remainder Theorem ka ek special case hai.

Remainder Theorem se: $p (x) = (x - a) \cdot q (x) + p (a)$

Forward direction (Agar $(x - a)$ ek factor hai, toh $p (a) = 0$ ):

Agar $(x - a)$ ek factor hai, toh $p (x)$ evenly divide hota hai koi remainder nahi
Iska matlab: $p (x) = (x - a) \cdot q (x) + 0$
$x = a$ substitute karne par: $p (a) = 0 + 0 = 0$ ✓

Reverse direction (Agar $p (a) = 0$ , toh $(x - a)$ ek factor hai):

Agar $p (a) = 0$ , toh Remainder Theorem se: $r = p (a) = 0$
Toh: $p (x) = (x - a) \cdot q (x) + 0 = (x - a) \cdot q (x)$
Iska matlab $(x - a)$ , $p (x)$ ko exactly divide karta hai, toh yeh ek factor hai ✓

Common Mistakes

Recall Ek 12 saal ke bacche ko explain karo

Imagine karo tumhare paas ek bada number hai jaise 17, aur tum usse 5 se divide karna chahte ho. Tumhe 3 milega remainder 2 ke saath, hai na?

Polynomials bhi isi tarah kaam karte hain! Agar tumhare paas $x^{3} - 4 x^{2} + 6 x - 5$ hai aur tum usse $(x - 2)$ se divide karo, toh koi quotient aur remainder milega.

Lekin yahan magic trick hai: tumhe actually long division karne ki zaroorat nahi. Bas apni polynomial mein 2 plug in karo (kyunki $x - 2 = 0$ jab $x = 2$ hota hai), aur jo bhi answer aaye — wahi tumhara remainder hai! Yeh math mein built-in shortcut calculator jaisa hai.

Aur ek bonus bhi hai: agar tumhara remainder exactly zero ho, iska matlab hai $(x - 2)$ tumhari polynomial mein perfectly fit hota hai — yeh ek factor hai, jaise 3, 12 ka factor hai kyunki woh evenly divide karta hai.

Key Applications

Polynomial long division ke bina quick remainder calculation
Potential roots test karke polynomials ke factors dhundna
Factoring karke polynomial equations solve karna
Polynomials mein unknown coefficients determine karna
Ek polynomial ki doosre se divisibility check karna

Connections

Polynomial Long Division — "slow way" jise yeh theorems shortcut karte hain
Rational Root Theorem — yeh identify karne mein madad karta hai ki $a$ ki kaun si values test karni hain
Fundamental Theorem of Algebra — roots (aur isliye factors) ke existence ki guarantee deta hai
Synthetic Division — Remainder Theorem use karne wali efficient computation technique
Factoring Polynomials — Factor Theorem main tool hai
Solving Polynomial Equations — roots dhundna = factors dhundna

Flashcards

#flashcards/maths

Remainder Theorem kya kehta hai? :: Jab polynomial $p (x)$ ko $(x - a)$ se divide kiya jaata hai, remainder $p (a)$ hota hai.

Factor Theorem state karo

(x - a)

p (x)

ka factor hai if and only if

p (a) = 0

p (x)

(2 x - 3)

se divide karte waqt remainder dhundhne ke liye tum kaun si value evaluate karte ho?

p (\frac{3}{2})

, kyunki

2 x - 3 = 0

jab

x = \frac{3}{2}

hota hai.

Agar

p (5) = 0

, toh tum kya conclude kar sakte ho?

(x - 5)

p (x)

ka factor hai.

Agar

(x + 4)

p (x)

ka factor hai, toh

p (- 4)

kya hai?

p (- 4) = 0

, kyunki

(x + 4) = (x - (- 4))

ek factor hai.

Linear polynomial se divide karne par remainder constant kyun hona chahiye?

Remainder ki degree divisor ki degree se kam honi chahiye. Kyunki ek linear polynomial ki degree 1 hoti hai, remainder ki degree 0 (constant) hoti hai ya woh zero hota hai.

Remainder Theorem 3 steps mein derive karo :: (1) $p (x) = (x - a) \cdot q (x) + r$ likhein; (2) $x = a$ substitute karke $p (a) = 0 \cdot q (a) + r$ pao; (3) Isliye $r = p (a)$ .

Remainder Theorem aur Factor Theorem mein kya relation hai?

Factor Theorem, Remainder Theorem ka ek special case hai jab remainder zero ke barabar hota hai.

Agar tum check karna chahte ho ki

(x - 7)

p (x) = x^{3} - 9 x + 2

ka factor hai ya nahi, toh tum kaun sa ek calculation perform karte ho?

p (7)

calculate karo. Agar yeh 0 ke barabar ho, toh

(x - 7)

ek factor hai; warna nahi.

(x - a)

p (x)

ka factor hone ka kya matlab hai?

p (x)

(x - a) \cdot q (x)

ke roop mein likha ja sakta hai kisi polynomial

q (x)

ke liye, zero remainder ke saath.