The Remainder Theorem and The Factor Theorem

Theorem ကိုမရွင္းျပခင္ Polynomial ဆိုတာကို ရွင္းျပခ်င္ပါတယ္။

Polynomial

ကိန္းရွင္(variable) တစ္ခုရဲ့ ထပ္ကိန္းမ်ား ပါ၀င္တဲ့ ကိန္းတန္း တစ္ခု (တနည္းေျပာရင္ function တစ္ခု) ကို polynomial လို႔ေခၚပါတယ္။ x ပါ၀င္တဲ့ ကိန္းတန္းတစ္ခုကို f(x),g(x), h(x),Q(x) စသည္ျဖင့္ သတ္မွတ္ႏိုင္ပါတယ္။ ဥပမာ ကိန္းတန္းမ်ားကို ၾကည့္ပါ။

Polynomial တစ္ခုရဲ့ အႀကီးဆံုးထပ္ကိန္းကို ၎ကိန္းတန္းရဲ့ order လို႔ သတ္မွတ္ႏိုင္ပါတယ္။ အထက္မွာ ေျပာခဲ့တဲ့ polynomial ေတြရဲ့ order ကို ေျပာရမယ္ဆိုရင္ f(x) က order 5၊ g(x) က order 4၊ h(x) က order 1 ျဖစ္ပါတယ္။

Remainder Theorem ဆိုတာက Polynomial of any order ကို polynomial of order 1 နဲ႔ စားတဲ့အခါ ရလာမယ့္ remainder (အၾကြင္း) ကို ရွာယူမွာ ျဖစ္ပါတယ္။

Remainder Theorem

If the polynomial f(x) is divided by (x-k) where k is a constant, the remainder is f(k).

f(x)÷ (x-k)=>Remainder=f(k))

Proof: Let Q(x) be the quotient and R be the remainder when f(x) is divided by (x-k).

Therefore f(x) = Q(x) (x-k) + R

f(k) = Q(k) (k-k) + R

f(k) = 0 +R and

f(k) = R.

Therefore the remainder theorem is satisfied.

Note:

Q(x) = quotient = စားလဒ္

f(x) = dividend = တည္ကိန္း

(x-k) = divisor = စားကိန္း (သို႔) စားေျခ

Dividend Divisor

= Quotient +

Remainder Divisor

Extension of Remainder Theorem

f(x)÷(x+k) => Remainder = f(-k)

f(x)÷(ax-b) => Remainder = f(b/a)

f(x)÷(ax+b) => Remainder = f(-b/a)

f(x)÷(p-qx) => Remainder = f(p/q)

f(x)÷(p+qx) => Remainder = f(-p/q)

f(x)÷ax => Remainder = f(0)

f(x)÷x => Remainder = f(0)

Example 3
Find the remainder when x³ + 4x² - 7x + 6 is divided by x - 1.

Let f(x) = x³ + 4x² - 7x + 6
f(1) = 1³ + 4 (1)² - 7 + 6
= 4

Example 4

Given that the expression 2x³ + 3px² - 4x + p has a remainder of 5 when divided by x + 2, find the value of p.

Let f(x) = 2x³ + 3px² - 4x + p

f (-2) = 2(-2)³ + 3(-2)²p - 4(-2) + p = 5

13p - 8 = 5

13p = 13

p = 1

Example 5

If the expression ax⁴ + bx³ - x² + 2x + 3 has remainder 4x + 3 when divided by x² + x - 2, find the value of a and b.

Let f(x) = ax⁴ + bx³ - x² + 2x + 3

x² + x - 2 = (x + 2)(x - 1)

f(-2) = a(-2)⁴ + b(-2)³ - (-2)² + 2(-2) + 3 = 4(-2) + 3

16a - 8b - 4 - 4 + 3 = -5

2a - b = 0 --------(1)

f(1) = a + b - 1 + 2 + 3 = 4(1) + 3

a + b = 3 --------(2)

(1) + (2) : 3a = 3

a = 1

when a = 1, b = 2.

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