求助4道因式分解(初2)1n(n+1)(n+2)(n+3)+12(x+1)(x+3)(x+5)(x+7)+153X2次+
2个回答

1.n(n+3)(n+1)(n+2)+1=(n^2+3n)(n^2+3n+2)+1

=(n^2+3n)^2+2(n^2+3n)+1=(n^2+3n+1)^2

2.(x^2+8x+7)(x^2+8x+15)+15=(x^2+8x+7)^2+8(x^2+8x+7)+15

=(x^2+8x+12)(x^2+8x+10)

=(x+2)(x+6)(x^2+8x+10)

3.x^2+3xy+2y^2+4x+5y+3=(x+y+a)(x+2y+b)

解出,a=1,b=3

所以,x^2+3xy+2y^2+4x+5y+3=(x+y+1)(x+2y+3)

这是一种待定系数法

4.12*A^((3N+1)-32*A^(2N-1)+24*A^(N-1)

=4A^(N-1)[3A^(2N+1)-8*A^N+6]