因式分解用特殊值法分解X^3+3X^2+3X+2（ax+by)^2+(bx-ay)^2(x+1)^4-2(x^2-1)+(x-1)^4-(x^2+3)^2x^4+x^2y^2+y^2

Question

因式分解
用特殊值法分解X^3+3X ^2+3X+2
（ax+by)^2+(bx-ay)^2
(x+1)^4-2(x^2-1)+(x-1)^4-(x^2+3)^2
x^4+x^2y^2+y^2

Answer 1

不太明白您的意思,是说 只是x³+3x²+3x+2用特殊值法么?
1）x³+3x²+3x+2
取 x=-2代入,原式=(-2)³+3×(-2)²+3×(-2)+2= -8+12-6+2=0
所以 x-(-2),即x+2是原式的一个因式
所以设x³+3x²+3x+2=(x+2)(x²+ax+b)=x³+ax²+bx+2x²+2ax+2b=x³+(a+2)x²+(b+2a)x+2b
比较系数,得 a+2=3,b+2a=3,2b=2
解得 a=1,b=1
所以x³+3x²+3x+2=(x+2)(x²+x+1)
2）(ax+by)^2+(bx-ay)^2
=a²x²+2abxy+b²y²+b²x²-2abxy+a²y²
= (a²x²+b²x²)+(a²y²+b²y²)
=x²(a²+b²)+y²(a²+b²)
=(x²+y²)(a²+b²)
3）(x+1)^4-2(x^2-1)+(x-1)^4-(x^2+3)^2
=(x²+2x+1)²-2(x²-1)+ (x²-2x+1)²-(x²+3)²
=[(x²+1)+2x]²- 2(x²-1)+ [(x²+1)-2x]²- (x²+3)²
=[(x²+1)²+4x(x²+1)+4x²]- 2(x²-1)+ [(x²+1)²-4x(x²+1)+4x²]-(x²+3)²
=2(x²+1)²+8x²-2(x²-1)-(x²+3)²
=2(x^4+2x²+1)+8x²-2x²+2-(x^4+6x²+9)
=2x^4+4x²+2+6x²+2-x^4-6x²-9
=x^4+4x²-5
=(x²+5)(x²-1)
=(x²+5)(x+1)(x-1)
4）x^4+x^2y^2+y^2
题目是不是写错了?
应是x^4+x²y²+y^4吧
x^4+x²y²+y^4
=( x^4+2x²y²+y^4)-x²y²
=(x²+y²)²-(xy)²
=(x²+xy+y²)(x²-xy+y²)