Question

因式分解(x^2+2x+2)^2+4(3x^2+7x+3)(2x^2+5x+1)

谢谢

Answer 1

设(x+1)^2=y,那么原式整理后
=[(x+1)^2+1]^2+4[3(x+1)^2+x][2(x+1)^2+(x-1)]
=(y+1)^2+4(3y+x)[2y+(x-1)]
=y^2+2y+1+4[6y^2+(5x-3)y+x(x-1)]
=25y^2+(20x-10)y+4x^2-4x+1
=(5y)^2+10(2x-1)y+(2x-1)^2
=[5y+(2x-1)]^2
=[5(x+1)^2+(2x-1)]^2
=(5x^2+12x+4)^2
=[(5x+2)(x+2)]^2
解法2
原式=(x^2+2x+2)^2+4[3(x^2+2x+2)+(x-3)][2(x^2+2x+2)+(x-3)]
=25(x^2+2x+2)^2+20(x-3)(x^2+2x+2)+4(x-3)^2
=[5(x^2+2x+2)+2(x-3)]^2
=(5x^2+12x+4)^2
=[(5x+2)(x+2)]^2