Question

因式分解x^4-2x^3-44x+7

Answer 1

用待定系数法就是要
设原式=(x²+ax+1)(x²+bx+7)或=(x²+ax-1)(x²+bx-7)
[因为首项可分解为x⁴=x²*x²
末项可分解为7=1*7或(-1)*(-7)]
然后再把括号打开再合并同类项可以得到：
x⁴+(a+b)x³+(ab+8)x²+(7a+b)x+7=x⁴-2x³-0x²-44x+7
或x⁴+(a+b)x³+(ab-8)x²-(7a+b)x+7=x⁴-2x³-0x²-44x+7
比较等式两边系数可得
a+b=-2
ab+8=0
7a+b=-44
或a+b=-2
ab-8=0
-7a-b=-44
解这两个方程组可得:
自己算吧，可能是这样的

Answer 2

没有简单的分解，你可能题目看错了，因为两实根是：（sqrt表示开方）
x1
=
1/2+1/2
sqrt(1-(10
2^(2/3))/(491+9
sqrt(3001))^(1/3)+(2
(491+9
sqrt(3001)))^(1/3))-1/2
sqrt(2+(10
2^(2/3))/(491+9
sqrt(3001))^(1/3)-(2
(491+9
sqrt(3001)))^(1/3)+90/sqrt(1-(10
2^(2/3))/(491+9
sqrt(3001))^(1/3)+(2
(491+9
sqrt(3001)))^(1/3)))
x
2=
1/2+1/2
sqrt(1-(10
2^(2/3))/(491+9
sqrt(3001))^(1/3)+(2
(491+9
sqrt(3001)))^(1/3))+1/2
sqrt(2+(10
2^(2/3))/(491+9
sqrt(3001))^(1/3)-(2
(491+9
sqrt(3001)))^(1/3)+90/sqrt(1-(10
2^(2/3))/(491+9
sqrt(3001))^(1/3)+(2
(491+9
sqrt(3001)))^(1/3)))
两虚根是
x3
=
1/2-1/2
sqrt(1-(10
2^(2/3))/(491+9
sqrt(3001))^(1/3)+(2
(491+9
sqrt(3001)))^(1/3))-1/2
sqrt(2+(10
2^(2/3))/(491+9
sqrt(3001))^(1/3)-(2
(491+9
sqrt(3001)))^(1/3)-90/sqrt(1-(10
2^(2/3))/(491+9
sqrt(3001))^(1/3)+(2
(491+9
sqrt(3001)))^(1/3)))
x4
=
1/2-1/2
sqrt(1-(10
2^(2/3))/(491+9
sqrt(3001))^(1/3)+(2
(491+9
sqrt(3001)))^(1/3))+1/2
sqrt(2+(10
2^(2/3))/(491+9
sqrt(3001))^(1/3)-(2
(491+9
sqrt(3001)))^(1/3)-90/sqrt(1-(10
2^(2/3))/(491+9
sqrt(3001))^(1/3)+(2
(491+9
sqrt(3001)))^(1/3)))