Phân tích đa thức sau thành nhân tử: 4(x + 5)(x + 6)(x + 10)(x + 12) – 3x^2

Ta có:

4(x+5)(x+6)(x+10)(x+12)−3x²

=4[(x+5)(x+12)][(x+6)(x+10)]−3x²

=4(x²+17x+60)(x²+16x+60)−3x²

=(2x²+34x+120)(2x²+32x+ 60)−3x²

=(2x²+33x+120 + x)(2x²+33x+ 60 – x)−3x²

=(2x²+33x+120)²–x²−3x²

=(2x²+33x+120)²–4x²

=(2x²+33x+120−2x)(2x²+33x+120+2x)

=(2x²+31x+120)(2x²+35x+120)

=(2x+15)(x+8)(2x²+35x+120).