(x2 + 2.x.3 + 32 - 1).(x2 + 2.x.4 + 16 - 1) - 24
=[(x+3)2 - 1]. [(x+4)2-1] -24
=(x+3+1)(x+3-1)(x+4+1)(x+4-1) - 24
=(x+4)(x+2)(x+5)(x-3) - 24
(x2+6x+8)(x2+8x+15)-24
<=>(x2+4x+2x+8)(x2+5x+3x+15)-24
<=> [x(x+4)+2(x+4)][x(x+5)+3(x+5)]-24
<=> (x+4)(x+2)(x+5)(x+3)-24
<=> (x+4)(x+3)(x+2)(x+5)-24
<=>(x2+7x+12)(x2+7x+10)
đặt t=x2+7x+11 ta có:
(t-1)(t+1)-24
<=> t2-1-24
<=>t2-25
<=>(t-5)(t+5)
thay t=x2+7x+11 vào ta có:
(x2+7x+11-5)(x2+7x+11+5)
<=>(x2+7x+6)(x2+7x+16)