Question

分解因式 初二试题

1）（x²+4x+8）²+3x（x²+4x+8）+2x²
2) (2x²-3x+1)²-22x²+33x-1
3) (6x-1)(2x-1)(3x-1)(x-1)+x²
4) ab(a+b)²-(a+b)²+1
5) x²+xy-6y²+x+13y-6

Answer 1

1）(x²+4x+8)²+3x(x²+4x+8)+2x²
=(x²+4x+8+x)(x²+4x+8+2x) (把x²+4x+8看做一个整体）
=(x²+5x+8)(x²+6x+8)
=(x²+5x+8)(x+2)(x+4)
2) (2x²-3x+1)²-22x²+33x-1
=(2x²-3x+1)²-11(2x²-3x+1)+10
=(2x²-3x+1-1)(2x²-3x+1-10) (这一步和第一题一样)
=(2x²-3x)(2x²-3x-9)
=x(2x-3)(x-3)(2x+3)
3) (6x-1)(2x-1)(3x-1)(x-1)+x²
=(6x-1)(x-1)(2x-1)(3x-1)+x²
=(6x²+1-7x)(6x²+1-5x)+x²
=(6x²+1)²-12(6x²+1)x+35x²+x²
=(6x²+1)²-12(6x²+1)x+36x²
=[(6x²+1)-6x]²
=(6x²-6x+1)²
4) ab(a+b)²-(a+b)²+1
=(ab-1)(a+b)²+1
=(ab-1)a²+(ab-1)b²+2ab(ab-1)+1
=(ab-1)a²+(ab-1)b²+2(ab)²-2ab+1
=(ab-1)a²+(ab-1)b²+(ab)²+(ab)²-2ab+1
=(ab-1)a²+(ab-1)b²+(ab)²+(ab²-2ab+1)
=(ab-1)a²+(ab-1)b²+(ab)²+(ab-1)²
=[(ab-1)a²+(ab)²]+[(ab-1)b²+(ab-1)²]
=a²[(ab-1)+b²]+(ab-1)*[(ab-1)+b²]
=(a²+ab-1)(b²+ab-1)
5) x²+xy-6y²+x+13y-6
=(x+3y)(x-2y)+(x+13y)-6 (这是关于x+3y和x-2y的式子)
=(x+3y-2)(x-2y+3)