Question

把下列各式分解因式(1) -4m³+16m²-26m

（2）x³-2x²y+xy² （3）（x²-3x）²-（3x-1）²

Answer 1

(1)
-4m³+16m²-26m
=-4m(m²-4m+26)
=-4m[(m-2)²+22]
=-4m(m-2+√22i)(m-2-√22i) /这道题要不就是抄错了，如果正确的，那么实数范围内无法因式分解，复数范围内可以。
(2)
x³-2x²y+xy²
=x(x²-2xy+y²)
=x(x-y)²
(3)
(x²-3x)²-(3x-1)²
=[(x²-3x)+(3x-1)][(x²-3x)-(3x-1)]
=(x²-1)(x²-6x+1)
=(x+1)(x-1)(x²-6x+1) /如果是实数范围内因式分解，继续：
=(x+1)(x-1)[(x-3)²-8]
=(x+1)(x-1)(x-3+2√2)(x-3-2√2)

Answer 2

(1)-4m³+16m²-16m
=-4m(m²-4m+4)
=-4m(m-2)²

（2）x³-2x²y+xy²
=x(x²-2xy+y²)
=x(x-y)²

（3）（x²-3x）²-（3x-1）²
=(x²-3x+3x-1)(x²-3x-3x+1)
=(x²-1)(x²-6x+1)
=(x+1)(x-2)(x²-6x+1)
=(x+1)(x-2)(x-3+2√2)(x-3-2√2)

Answer 3

x³-2x²y+xy²=x(x²-2xy+x²)=x(x-y)²

（x²-3x）²-（3x-1）²
=（x²）²-6x*x²+9x²-9x²+6x-1
=（x²）²-6x*x²+6x-1
=（x²）²-1-(6x*x²-6x)
=(x²+1)(x²-1)-6x(x²-1)
=(x²-1)(x²-6x+1)
=(x+1)(x-1)(x²-6x+1)