Question

求解两道因式分解

x³-11x²+31x-21

x³-4xy²-2x²y+8y过程详细点

Answer 1

原式=(x³-1)-(11x³-31x+20)
=(x-1)(x²+x+1)-(11x-20)(x-1)
=(x-1)(x²+x+1-11x+20)
=(x-1)(x²-10x+21)
=(x-1)(x-3)(x-7)

x³-2x²y-4xy-8y-8
=(x³-8)-(2x²y+4xy+8y)
=(x³-8)-2y(x²+2x+4)
=(x-2)(x²+2x+4)-2y(x²+2x+4)
=(x²+2x+4)(x-2y-2)

Answer 2

x³-11x²+31x-21
=x³-1-(11x²-31x+20)
=(x-1)(x²+x+1)-(11x-20)(x-1)
=(x-1)(x²+x+1-11x+20)
=(x-1)(x²-10x+21)
=(x-1)(x-3)(x-7)

最后是y³
x³-4xy²-2x²y+8y
=x(x²-4y²)-2y(x²-4y²)
=(x-2y)(x²-4y²)
=(x-2y)(x+2y)(x-2y)
=(x+2y)(x-2y)²

Answer 3

x³-11x²+31x-21
=x³-x²-10x²+10x+21x-21
=x²(x-1)-10x(x-1)+21(x-1)
=(x-1)(x²-10x+21)
=(x-1)(x-3)(x-7)

x³-4xy²-2x²y+8y³
=x²(x-2y)-4y²(x-2y)
=(x-2y)(x²-4y²)
=(x-2y)²(x+2y)