數學恆等式與因式分解
回答 (4)
2011-12-28 2:08 am
✔ 最佳答案
1 (x - 2)(x - 3) = x^2 + Bx + Cx^2 - 5x + 6 = x^2 + Bx + C
比較兩邊係數﹐B = -5, C = 6
2012-01-20 12:07 am
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(x-2)(x-3) ≡ x²+ Bx + C
x(x-3) - 2(x-3) ≡ x²+ Bx + C
x²- 3x - (2x-6) ≡ x²+ Bx + C
x²- 3x - 2x + 6 ≡ x²+ Bx + C
x²- 5x + 6 ≡ x²+ Bx + C
-5x = Bx
B = -5
6 = C
C = 6
答：B 和 C 的值分別為 -5 和 6 。
(x-2)(x-3) ≡ x²+ Bx + C
x(x-3) - 2(x-3) ≡ x²+ Bx + C
x²- 3x - (2x-6) ≡ x²+ Bx + C
x²- 3x - 2x + 6 ≡ x²+ Bx + C
x²- 5x + 6 ≡ x²+ Bx + C
-5x = Bx
B = -5
6 = C
C = 6
答：B 和 C 的值分別為 -5 和 6 。
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2011-12-28 3:44 am
(x-2)(x-3)=x^2+Bx+C
左=x^2-3x-2x+6
左=x^2-5x+6
B=-5 C=6
左=x^2-3x-2x+6
左=x^2-5x+6
B=-5 C=6
2011-12-28 2:52 am
(x - 2)(x - 3) = x^2 + Bx + C
put x = 2,
0 = 4 + 2B + C
put x = 3,
0 = 9 + 3B + C
solving for two equations,
B = -5 and C = 6
put x = 2,
0 = 4 + 2B + C
put x = 3,
0 = 9 + 3B + C
solving for two equations,
B = -5 and C = 6
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