中4續多項式(急~!!!!!!!!!!!!!!!!!!!)

1.
a)
P(x) 可被 (x + 3) 整除，故此 P(-3) = 0
(-3)³ + a(-3)² + b(-3) - 6 = 0
3a - b = 11 ... [1]

當 P(x) 除以 x - 2 時，餘數為 20，故此 P(2) = 20
(2)³ + a(2)² + b(2) - 6 = 20
2a + b = 9 ... [2]

[1] + [2] :
5a = 20
a = 4

將 a = 4 代入戋 [2] :
2(4) + b = 9
b = 1

b)
用長除法，P(x)
= x³ + 4x² + x - 6
= (x + 3)(x² + x - 2)
= (x + 3)(x + 2)(x - 1)


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2)
a)
P(x) 可被 (x + 5) 整除，故此 P(-5) = 0
2(-5)³ + a(-5)² + b(-5) - 15 = 0
10a - 2b = 106 ... [1]

當 P(x) 除以 2x - 1 時，餘數為 -45/2，故此 P(-1/2) = -45/2
2(-1/2)³ + a(-1/2)² + b(-1/2) - 15 = -45/2
a - 2b = -29 ... [2]

[1] - [2] :
9a = 135
a = 15

將 a = 15 代入 [2] :
15 - 2b = -29
b = 22

b)
用長除法，P(x)
= 2x³ + 15x² + 22x - 15
= (x + 5)(2x² + 5x - 3)
= (x + 5)(2x - 1)(x + 3)


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3)
a)
P(x) 可被 (x - 6) 整除，故此 P(6) = 0
3(6)³ + a(6)² + b(6) + 12 = 0
12a + 2b = -220 ... [1]

當 P(x) 除以 2x - 3 時，餘數為 99/8，故此 P(3/2) = 99/8
3(3/2)³ + a(3/2)² + b(3/2) + 12 = 99/8
3a + 2b = -13 ... [2]

[1] - [2] :
9a = -207
a = -23

將 a = -23 代入 [2] :
3(-23) + 2b = -13
b = 28

b)
用長除法，P(x)
= 3x³ - 23x² + 28x + 12
= (x - 6)(3x² - 5x - 2)
= (x - 6)(3x + 1)(x - 2)


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4)
P(x) 可被 (2x + 5) 整除，故此 P(-5/2) =0
a(-5/2)³ + (-5/2)² + b(-5/2) - 15 = 0
25a + 4b = -14 ... [1]

當 P(x) 除以 3x - 1 時，餘數為 -544/27，故此 P(1/3) = -544/27
a(1/3)³ + (1/3)² + b(1/3) - 15 = -544/27
25a + 225b = -3550 ... [2]

[2] - [1]
221b = -3536
b = -16

將 b = -16 代入 [1] :
25a + 4(-16) = -14
a = 2

b)
用長除法，P(x)
= 2x³ + x² - 16x - 15
= (2x + 5)(x² - 2x - 3)
= (2x + 5)(x - 3)(x + 1)


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5)
P(x) 可被 (5x - 2) 整除，故此 P(2/5) =0
a(2/5)³ + b(2/5)² - 5(2/5) + 6 = 0
2a + 5b = -125 ... [1]

當 P(x) 除以 3x + 2 時，餘數為 -176/27，故此 P(-2/3) = -176/27
a(-2/3)³ + b(-2/3)² - 5(-2/3) + 6 = 0
2a - 3b = 107 ... [2]

[1] - [2] :
8b = -232
b = -29

將 b = -29 代入 [2] :
2a - 3(-29) = 107
a = 10

b)
用長除法，P(x)
= 10x³ - 29x² - 5x + 6
= (5x - 2)(2x² - 5x - 3)
= (5x - 2)(2x + 1)(x - 3)