國2上數學

1.
(A - B) * C
= [(4x² - x) - (5x² + 2)] * (x² - 4x - 3)
= (4x² - x - 5x² - 2) * (x² - 4x - 3)
= (-x² - x - 2) * (x² - 4x - 3)
= -x²(x² - 4x - 3) - x(x² - 4x - 3) - 2(x² - 4x - 3)
= -x⁴ + 4x³ + 3x² - x³ + 4x² + 3x - 2x² + 8x + 6
= -x⁴ + 3x³ + 5x² + 11x + 6


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2.
2x³ + 3x² + kx - 3 = (2x + 3)(ax² + b)
2x³ + 3x² + kx - 3 = 2ax³ + 3ax² + 2bx + 3b

比較常數項:
3b = -3
b = -1

比較 x 項:
k = 2b
k = 2*(-1)
k = -2


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3.
[(17x² - 3x + 4) - (ax² + bx + c)] ÷ (5x + 6) = 2x+ 1
(17x² - 3x + 4) - (ax² + bx + c) = (2x + 1)(5x +6)
(17 - a)x² - (3 + b)x + (4 - c) = 10x² + 17x + 6

比較 x²項:
17 - a = 10
a = 7

比較 x 項:
-(3 + b) = 17
-3 - b = 17
b = -20

比較常數項:
4 - c = 6
c = -2

a - b - c = 7 - (-20) - (-2)
a - b - c = 29


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4.
題目不完整。


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5.
(1)
（此題可用長除法計算。）

設商式為 x² + ax + b。

(x⁴ - x³ + 2x² - mx + n) ÷ (x² + x - 1) = x² + ax + b ..... 餘-2x + 3
x⁴ - x³ + 2x² - mx + n = (x² + x - 1)(x² + ax + b) + (-2x + 3)
x⁴ - x³ + 2x² - mx + n = x⁴ + (a + 1)x³ + (a + b - 1)x² + (-a + b)x - b - 2x +3
x⁴ - x³ + 2x² - mx + n = x⁴ + (a + 1)x³ + (a + b - 1)x² + (-a + b - 2)x + (-b +3)

比較 x³項:
-1 = a + 1
a = -2

比較 x²項：
2 = a + b - 1
-2 + b - 1 = 2
b = 5

商式 =x² - 2x + 5

(2)
（此題可用長除法計算。）

x⁴ - x³ + 2x² - mx + n = x⁴ + (a + 1)x³ + (a + b - 1)x² + (-a + b - 2)x + (-b +3)

比較 x 項:
-m = -a + b - 2
-m = -(-2) + 5 - 2
m = -5

比較常數項:
n = -b + 3
n = -5 + 3
n = -2


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6.
（此題可用長除法計算。）

設q(x) = 2x² + ax + b 及 r(x) = mx + n

(2x⁴ + 3x³ + x - 1) ÷ (x² - 1) = 2x² + ax + b ......餘 mx + n
2x⁴ + 3x³ + x - 1 = (x² - 1)(2x² + ax + b) + (mx +n)
2x⁴ + 3x³ + x - 1 = (x² - 1)(2x² + ax + b) + (mx +n)
2x⁴ + 3x³ + x - 1 = 2x⁴ + ax³ + (b - 2)x² - ax - b + mx + n
2x⁴ + 3x³ + x - 1 = 2x⁴ + ax³ + (b - 2)x² + (-a + m)x + (-b + n)

比較 x³項:
a = 3

比較 x²項:
b - 2 = 0
b = 2

比較 x 項:
-a + m = 1
-3 + m = 1
m = 4

比較常數項:
-b + n = -1
-2 + n = -1
n = 1

q(x) = 2x² + 3x + 2
r(x) = 4x + 1

q(1) + r(1)
= (2 + 3 + 2) + (4 + 1)
= 7 + 5
= 12

1. [(4x^2-x)-(5x^2+2)] * (x^2-4x-3)

=>(-x^2-x-2) * (x^2-4x-3)

=>-x^4 + 4x^3 + 3x^2 - x^3 + 4x^2 + 3x - 2x^2 + 8x + 6

=>-x^4 + 3x^3 + 5x^2 + 11x +6

2. 這表示 "2X+3" 為 "2X^3 + 3X^2 + KX - 3" 的因式

(2x+3) * (ax^2+b) = 2X^3 + 3X^2 + KX - 3

=> 2ax^3 + 2bx + 3ax^2 + 3b

=> b=-1 =====> K=2b=-2