→(x+5)(x+9)(x+3)(x+7)−33
=(x+3)(x+9)(x+7)(x+5)−33
=[x(x+9)+3(x+9)][x(x+5)+7(x+5)]−33
=[x2+9x+3x+27][x2+5x+7x+35]−33
=[x2+12x+27][x2+12x+35]−33
=[a+27][a+35]−33
=a(a+35)+27(a+35)−33
=a2+35a+27a+945−33
=a2+62a+912
=a2+24a+38a+912
=a(a+24)+38(a+24)
=(a+24)(a+38)=[x2+12x+24][x2+12x+38]