1)
已知f(x) = 2x^3 - x^2 - 7x + 6。求f(3/2),由此,因式分解f(x)
2)
已知f(x) = 3x^3 - 10x^2 + x + 6。求f(-2/3),由此,因式分解f(x)
3)
已知H(x)= 2x^3 + px^2 - 4qx -15 可被x - 5 和2x+1整除,
a.求p和q的值
b.由此,求多項式的其他因式
4)
已知f(x) = f(x) = x^3 + 5x^2 - 2ax - b , g(x) = ax^3 - 7x^2 - 6x +3b。
若x + 3是f(x)和g(x)的公因式,
a.求a和b的值
b.分析x - 1 是否 (i) f(x)+g(x) , (ii) f(x) - g(x) 的因式
thank you!!
多項式,,希望可以幫忙..thanks!!
2007-03-29 4:51 am
回答 (2)
2007-03-29 6:16 am
✔ 最佳答案
(1)圖片參考：http://i117.photobucket.com/albums/o61/billy_hywung/Maths/Crazypoly1.jpg
(2)
圖片參考：http://i117.photobucket.com/albums/o61/billy_hywung/Maths/Crazypoly2.jpg
(3)
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(4)
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參考: My Maths knowledge
2007-03-29 5:52 am
1. f(3/2) = 2 (27/8) - (9/4) - 7(3/2) + 6 = 0
So by factor theorem, (2x - 3) is a factor of f
Hence, by factorization, f(x) = (2x - 3)(x^2 - x - 2) = (2x - 3)(x - 2)(x + 1)
2. f(-2/3) = 3 (-8/27) - 10 (4/9) + (-2/3) + 6 = 0
So by factor theorem, (3x + 2) is a factor of f
Hence, by factorization, f(x) = (3x + 2)(x^2 - 4x + 3) = (3x + 2)(x - 1)(x - 3)
3. By factor theorem, we have H(5) = H(-1/2) = 0
So we have
25p -20q + 235 = 0 (i.e. 5p - 4q = - 47)
p/4 + 2q - 61/4 = 0 (i.e. p + 8q = 61)
Solving gives p = -3, q = 8
Hence f(x)= (x - 5)(2x + 1)(x + c) for some c, by equating the constant term, we get
c = -3
So f(x)= (x - 5)(2x + 1)(x - 3)
4. By factor theorem, we have f(-3) = g(-3) = 0
which gives
6a - b + 18 = 0
-27a + 3b - 45 = 0
Solving gives a = 1, b = 24
Since f(1) + g(1) and f(1) - g(1) are not zero, so by factor theorem, x - 1 are not
factors of f + g and f - g
So by factor theorem, (2x - 3) is a factor of f
Hence, by factorization, f(x) = (2x - 3)(x^2 - x - 2) = (2x - 3)(x - 2)(x + 1)
2. f(-2/3) = 3 (-8/27) - 10 (4/9) + (-2/3) + 6 = 0
So by factor theorem, (3x + 2) is a factor of f
Hence, by factorization, f(x) = (3x + 2)(x^2 - 4x + 3) = (3x + 2)(x - 1)(x - 3)
3. By factor theorem, we have H(5) = H(-1/2) = 0
So we have
25p -20q + 235 = 0 (i.e. 5p - 4q = - 47)
p/4 + 2q - 61/4 = 0 (i.e. p + 8q = 61)
Solving gives p = -3, q = 8
Hence f(x)= (x - 5)(2x + 1)(x + c) for some c, by equating the constant term, we get
c = -3
So f(x)= (x - 5)(2x + 1)(x - 3)
4. By factor theorem, we have f(-3) = g(-3) = 0
which gives
6a - b + 18 = 0
-27a + 3b - 45 = 0
Solving gives a = 1, b = 24
Since f(1) + g(1) and f(1) - g(1) are not zero, so by factor theorem, x - 1 are not
factors of f + g and f - g
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