F.4 maths

Let f ( x ) = x^3 + x^2 - 4x + 2,

f ( k ) = k^2 ( k - 1 ) ( given )

k^3 + k^2 - 4k + 2 = k^2 ( k - 1 )

k^3 + k^2 - 4k + 2 = k^3 - k^2

2k^2 - 4k + 2 = 0

k^2 - 2k + 1 = 0

( k - 1 )^2 = 0

k = 1

2a) Let f ( x ) = x^3 - x^2 - 4x - 2,

f ( - 1 ) = ( - 1 )^3 - ( - 1 )^2 - 4 ( - 1 ) - 2

= - 1 - 1 + 4 - 2

= 0

So x + 1 is a factor of f ( x ).

b) x^3 - x^2 - 4x - 2 = 0

( x + 1 )( x^2 - 2x - 2 ) = 0

( x + 1 )( x^2 - 2x + 1 - 3 ) = 0

( x + 1 )[ ( x - 1 )^2 - 3 ] = 0

x + 1 = 0 or ( x - 1 )^2 = 3

x = -1 or 1 + sqr 3 or 1 - sqr 3

1. k=1
2. x+1=0 -- > x=-1 --> (-1)^3 - (-1)^2 - 4(-1) -2 = 0 Therefore x+1 is a root
3. (x+1)(x+1)(x-2)

2008-02-16 13:21:08 補充：
3 is wrong sor