Question

因式分解：(x的平方-3x-3)(x的平方-3x+4)-8 x(x-1)(x+1)(x+2)-24

Answer 1

(x的平方-3x-3)(x的平方-3x+4)-8
=(x²-3x)²+(x²-3x)-20
=(x²-3x+5)(x²-3x-4)
=(x²-3x+5)(x-4)(x+1)

x(x-1)(x+1)(x+2)-24
=[x(x+1)][(x-1)(x+2)]-24
=(x²+x)(x²+x-2)-24
=(x²+x)²-2(x²+x)-24
=(x²+x-6)(x²+x+4)
=(x+3)(x-2)(x²+x+4)

Answer 2

(x的平方-3x-3)(x的平方-3x+4)-8
=(x²-3x)²+4(x²-3x)-3(x²-3x)-12-8
=(x²-3x)²+(x²-3x)-20
=(x²-3x+5)(x²-3x-4)
=(x+1)(x-4)(x²-3x+5)

x(x-1)(x+1)(x+2)-24
=x(x²-1)(x+2)-24
=x(x³+2x²-x-2)-24
=x*x³+2x³-x²-2x-24
=x*x³+2x³-3x²+2x²-2x-24
=x²(x²+2x-3)+2(x²-x-12)
=x²(x+3)(x-1)+2(x-4)(x+3)
=(x+3)(x³-x²+2x-8)