Polynomials

Shinrei

Find the value of a for which ( x-2 )is a factor of 3x(cube) + ax(square)+ x -2
Hence, show that for this value of a, the cubic equation 3x(cube) + ax(square)+ x -2 = 0 has only 1 real root.

hamilan911

x=2 is one of the root
3(2^3) + a(2^2) + 2 - 2 = 0
24 + 4a = 0
a = -6
用long division
3x^3 - 6x^2 + x - 2 = (x-2)(3x^2+1)
3x^2 + 1 > 0 for all values of x
所以 3x^3 - 6x^2 + x - 2 = 0 只有一个root.

Shinrei

原帖由 hamilan911 于 1-6-2007 11:52 PM 发表
x=2 is one of the root
3(2^3) + a(2^2) + 2 - 2 = 0
24 + 4a = 0
a = -6
用long division
3x^3 - 6x^2 + x - 2 = (x-2)(3x^2+1)
3x^2 + 1 > 0 for all values of x
所以 3x^3 - 6x^2 + x - 2 = 0 只 ...

3x^2 + 1 > 0 for all values of x
这个不是代表 have 2 different roots？
我自己做做下做到以下的...你帮我看看对吗...
找到了a = -6 之后，我用synthetic division,

找到的quotient，3x^2 + 1
然后，f(x)= (x-2)(3x^2 + 1)
given that f(x) have 1 root,therefore f(x)=0
(x-2)(3x^2 + 1)= 0
(x-2)=0  ,  (3x^2 + 1)=0
   x =2  ,   x = [-1/3] << square root
             * no solution

x = 2 (shown)

hamilan911

原帖由 Shinrei 于 2-6-2007 12:07 AM 发表
3x^2 + 1 > 0 for all values of x
这个不是代表 have 2 different roots？
我自己做做下做到以下的...你帮我看看对吗...
找到了a = -6 之后，我用synthetic division,
找到的quotient，3x^2 + 1然 ...

3x^2 + 1 > 0 代表 there is no real solution for  3x^2 + 1 = 0
那个2 different roots 的是 b^2 -  4ac > 0
要用 b^2 - 4ac来证明也行
b^2 - 4ac = 0 - 4(1)(3) = -12 < 0
b^2 - 4ac < 0 shows that no solution for 3x^2 + 1 = 0

Shinrei

原帖由 hamilan911 于 2-6-2007 10:44 AM 发表

3x^2 + 1 > 0 代表 there is no real solution for  3x^2 + 1 = 0
那个2 different roots 的是 b^2 -  4ac > 0
要用 b^2 - 4ac来证明也行
b^2 - 4ac = 0 - 4(1)(3) = -12 < 0
b^2 - 4ac < 0  ...

明白了！谢谢你