因式分解(6x﹢7)的平方乘以(3x+4)乘以(x+1)减去6

2024-11-29 10:53:23
推荐回答(2个)
回答1:

(6x+7)^2(3x+4)(x+1)-6
=(36x²+84x+49)(3x²+7x+4)-6
=(36x²+84x+48+1)(3x²+7x+4)-6
=[12(3x²+7x+4)+1](3x²+7x+4)-6
=12(3x²+7x+4)²+(3x²+7x+4)-6
=[3(3x²+7x+4)-2][3(3x²+7x+4)+3]
=[3(3x²+7x+4)-2][4(3x²+7x+4)+3]
=[9x²+21x+12-2][12x²+28x+16+3]
=(9x²+21x+10)(12x²+28x+19)
=(3x+2)(3x+5)(12x²+28x+19)

回答2:

原式=(36x²+84x+49)(3x²+7x+4)-6
令a=3x²+7x+4
则36x²+84x+49=12a+1
原式=a(12a+1)-6
=12a²+a-6
=(3a-2)(4a+3)
=(9x²+21x+10)(12x²+28x+19)
=(3x+2)(3x+5)(12x²+28x+19)