factorization urgent!!

let f(k) = 2k^4+16k^3+47k^2+60k+28
代唔同既 k 值, 如 k = 1, k = -1, k = -2, 睇下 f(k) 等唔等於零
試到 f(-2) = 0, 即係話 (k + 2) 係其中一個 factor
長除法, 得 f(k) = (k + 2)(2k^3 + 12k^2 + 23k + 14)

再做多次, 搵 (2k^3 + 12k^2 + 23k + 14) 既 factor
應該搵到 (k + 2) 係 factor,
所以 2k^3 + 12k^2 + 23k + 14 = (k + 2)(2k^2 + 8k + 7)

即係
f(k)
= (k + 2)(k + 2)(2k^2 + 8k + 7)
= 2(k + 2)(k + 2)(k^2 + 4k + 7/2)

最後分解埋 (k^2 + 4k + 7/2)
用 quadratic formula, 搵到兩個根 a, b
a = (-4 + sqrt(2))/2, b = (-4 - sqrt(2))/2
即係話 (k^2 + 4k + 7/2) = (k - a)(k - b)

結合返埋,
2k^4+16k^3+47k^2+60k+28
= 2(k + 2)(k + 2)(k - a)(k - b)

a = (-4 + sqrt(2))/2, b = (-4 - sqrt(2))/2

做就係咁做,唔知你明唔明,
都係用多項式最基本既概念,

2007-08-07 21:56:38 補充：
同上面位人兄做法一樣,不過我一步一步做埋比你睇

設 f ( x ) = 2k^4+16k^3+47k^2+60k+28
f (-2) = 32+(-128)+188+(-120)+28
=0
.'.(x+2)是2k^4+16k^3+47k^2+60k+28其中一個因式
2k^4+16k^3+47k^2+60k+28
=(x+2)(2x^3+12x^2+23x+14)
=(x+2)(x+2)(2x^2+8x+7)(如果唔係實數以內嘅話)
=2[x+ ((4-√2)/2)] [x+ ((4+√2)/2)] (x+2)^2(如果係實數以內嘅話)