Question

高一分解因式，求答案，急！！！

x³-11x²+31x-21

（拆项法）x³-3x²+4
第一题次数分别是3、2、1】第二题是3.2

Answer 1

1、原式=X3一10X2+21X－X2+10X－21=(X－3)(X－7)(X－1)
2、原式=X3－2X2(2倍X平方)－X2+4=X2(X－2)－(2－X)(2+X)=(X－2)(X2+X+2)

Answer 2

x³-11x²+31x-21

=x³-11x²+10x+21x-21
=x(x²-11x+10)+21(x-1)
=(x-1)[x(x-10)+21]
=(x-1)(x²-10x+21)
=(x-1)(x-3)(x-7)
x³-3x²+4
=x³-3x²+2x-2x+4
=x(x²-3x+2)-2(x-2)
=(x-2)[x(x-1)-2]
=(x-2)(x²-x-2)
=(x-2)²(x+1)

Answer 3

x³-11x²+31x-21

=x³-11x²+10x+21x-21
=x(x²-11x+10)+21(x-1)
=x(x-1)(x-10)+21(x-1)
=(x-1)[x(x-10)+21]
=(x-1)(x²-10x+21)
=(x-1)(x-3)(x-7)

x³-3x²+4
=x³-2x²-x²+4
=x²(x-2)-(x-2)(x+2)
=(x-2)(x²-x-2)
=(x-2)²(x+1)