Remainder Theorem ကိုအသံုးခ်တြက္ေသာ ပုစာၦမ်ား(3)

Example 10

The remainder when x4 + 3x2 - 2x + 2 is divided by x + a is the square of the remainder when x2 - 3 is divided by x + a, Calculate the possible values of a.
x4 + 3x2 - 2x + 2 ကို x + a ႏွင့္စားေသာအခါ ရတဲ့ remainder သည္ x2 - 3 ကို x + a ႏွင့္ စားေသာအခါ ရတဲ့ remainder၏ ႏွစ္ထပ္ႏွင့္ ညီလွ်င္ a ၏ တန္ဖိုးမ်ားကို ရွာပါ။

http://i627.photobucket.com/albums/tt352/Thu-Rein/template/th_bluearrow.gif Let f(x) = x4 + 3x2 - 2x + 2 and g(x) = x2 - 3
By the problem,
f(-a) = [g(-a)]2
(-a)4 + 3(-a)2 - 2(-a) + 2 =[(-a)2 - 3]2
a4 + 3a2 + 2a + 2 = a4 - 6a2 + 9
9a2 + 2a - 7 = 0
(a + 1) (9a - 7) = 0
a = -1 (or) a=7/9

Example 11

The expression ax3 - x2 + bx - 1 leaves the remainder of -33 and 77 when divided by x + 2 and x - 3 respectively. Find the values of a and b and the remainder when divided by x - 2.
ax3 - x2 + bx - 1 ကို x + 2 နဲ႔စားတဲ့အခါ -33 ရၿပီး x - 3 နဲ႔စားတဲ့အခါ 77 ရပါတယ္။ a ႏွင့္ b ကိုရွာပါ။ x - 3 ႏွင့္စားလို႔ရတဲ့ remainder ကိုလည္းရွာေပးပါ။

http://i627.photobucket.com/albums/tt352/Thu-Rein/template/th_bluearrow.gif Let f(x) = ax3 - x2 + bx - 1.
By the problem,
f(-2) = -33
a(-2)3 - (-2)2 + b(-2) - 1 = -33
4a + b = -33 -------------(1)

f(3) = 77
a(3)3 - (3)2 + b(3) - 1 = 77
9a + b = 29 -------------(2)

equation (2) - equation (1),we get
5a = 15 and a = 3

Substitute a = 3 in equation (1)
4(3) + b = 14
b = 2