Remainder Theorem ကိုအသံုးခ်တြက္ေသာ ပုစာၦမ်ား(1)

Example 6

The polynomial x³ + ax² + bx - 3 leaves a remainder of 27 when divided by x - 2 and a remainder of 3 when divided by x + 1.Calculate the remainder when the polynomial is divided by x - 1

x³ + ax² + bx - 3 ကို x - 2 နဲ႔စားရင္ remainder 27 ရတယ္။ x + 1 နဲ႔ စားရင္ေတာ့ remainder 3 ရတယ္။ x - 1 နဲ႔ စားလို႔ရတဲ့ remainder ကိုရွာေပးပါ။

ရွင္းလင္းခ်က္။ ။ ေပးထားတဲ့ polynomial ကို f(x)လို႔ထားမယ္။ f(x) ကို x - 1 နဲ႔စားရင္ remainder f(1) ေပါ့။ ဒါေပမယ့္ unknown constant a နဲ႔ b ကိုသိရမယ္။ အဲဒါကို ေပးခ်က္ႏွစ္ခ်က္ကေန အရင္ရွာမယ္။ a နဲ႔ b ကို သိရင္ ဆက္ရွာ လို႔ ရၿပီေပါ့။

Let f(x) = x³ + ax² + bx - 3, by the problem

f(2) = 27

2³ + a(2²) + 2b - 3 = 27

2a + b = 11 -------------(1)

f(-1) = 3

(-1)³ + a(-1)² - b - 3 = 27

a - b = 31 ----------------(2)

equation (1) + equation (2) => 3a = 42 and a = 14

Substitute a = 14 in equation (2),

14 - b = 31

b = -17

Therefore f(x) = x³ + 14x² - 17x - 3

f(x) ÷ (x - 1) => remainder = f(1)

f(1) = 1³ + 14(1²) - 17(1) - 3 = - 5

Example 7

The expression x³ - 7x + 6 and x³ - x² - 4x +24 have the same remainder when divided by x + p. Find the possible values of p.
ကိန္းတန္းႏွစ္ခု ေပးထားပါတယ္။ ဒါေၾကာင့္ ပထမကိန္းတန္းကို f(x) လို႔ထားမယ္၊ ဒုတိယကိန္းတန္းကို g(x) လို႔ ထားမယ္။ ႏွစ္ခုလံုးကို x + p နဲ႔ စားတဲ့အခါ remainder တူတယ္။ p ရဲ့တန္ဖိုး ရွာေပးရမယ္။

Let f(x) = x³ - 7x + 6 and g(x) = x³ - x² - 4x +24
By the problem, f(-p) = g(-p)
p³ - 7p + 6 = p³ - p² - 4p +24
p² - 3p - 18 = 0
(p - 6) (p + 3) = 0
p = 6 or p = -3