✔ 最佳答案
(x^2+Ax+B)^2
=x^4+2Ax^3+(A^2+2B)x^2+2ABx+B^2
Comparing coefficient for x^3,
2A=-8 => A=-4
Comparing coefficient for x^2,
(A^2+2B)=22 so 16+2B=22 => B=3
So (x^2-4x+3)^2=x^4-8x^3+22x^2-24x+9
......
2009-06-14 01:44:51 補充：
Expressing a quartic polynomial f(x) into sum of 2 squares is not always possible because it means the quartic polynomial must be positive, so no real root for f(x)=0. So I guess your answer is for specific polynomials only.
2009-06-14 01:45:01 補充：
2 x^4+18 x^3+41 x^2-18 x+2
=(x^2+ax+b)^2+(x^2+cx+d)^2
=2x^4+2(a+c)x^3+(a^2+c^2+2b+2d)x^2+2(ab+cd)x+(b^2+d^2)=0
2009-06-14 01:45:11 補充：
Comparing terms,
2(a+c)=18...(1)
a^2+c^2+2b+2d=41...(2)
2(ab+cd)=-18...(3)
b^2+d^2=2...(4)
2009-06-14 01:45:17 補充：
Since it is specific polynomial, I have to make some guess. Looking at (4) I would guess b=+/-1 and d=+/-1
Put these combinations into (3) will always contradicts (1) except when b=-1 and d=-1.
Sub into (2) gives a^2+c^2=45
Combine with (1) gives a,c=6,3 or 3,6
So it is (x^2+3x-1)^2 + (x^2+6x-1)^2
2009-06-14 01:47:19 補充：
The senstence should read:
Put these combinations into (3) will always contradicts (1) and/or (2) except when b=-1 and d=-1.
