Question

因式分解(x^2+4x+3)(x^2+12x+35)+15

Answer 1

(x^2+4x+3)(x^2+12x+35)+15
=(x+1)(x+3)(x+5)(x+7)+15
=[(x+1)(x+7)][(x+3)(x+5)]+15
=(x^2+8x+7)(x^2+8x+15)+15
令K=x^2+8x+7
则
上式
=K(K+8)+15
=K^2+8K+15
=(K+3)(K+5)
=(x^2+8x+7+3)(x^2+8x+7+5)
=(x^2+8x+10)(x^2+8x+12)
=(x+2)(x+6)(x^2+8x+10)

Answer 2

(x^2+4x+3)(x^2+12x+35)+15

=(x+1)(x+3)(x+5)(x+7)+15
=[(x+1)(x+7)][(x+3)(x+5)]+15
=(x²+8x+7)(x²+8x+15)+15
令x²+8x=t
原式=(t+7)(x+15)+15
=t²+22t+120
=(t+12)(t+10)
=(x²+8x+12)(x²+8x+10)
=(x+2)(x+6)(x²+8x+10)

Answer 3

原式=(x+1)(x+3)(x+5)(x+7)+15
=[(x+1)(x+7)][(x+3)(x+5)]+15
=[(x²+8x)+7][(x²+8x)+15]+15
=(x²+8x)²+22(x²+8x)+105+15

=(x²+8x)²+22(x²+8x)+120
=(x²+8x+10)(x²+8x+12)
=(x²+8x+10)(x+2)(x+6)

Answer 4

式子中^2表示平方
(x^2+4x+3)(x^2+12x+35)+15
=(x+1)(x+3)(x+5)(x+7)+15
=(x+1)(x+7)(x+5)(x+3)+15
=(x^2+8x+7)(x^2+8x+15)+15
=(x^2+8x)^2+22(x^2+8x)+105+15
=(x^2+8x)^2+22(x^2+8x)+120
=(x^2+8x+10)(x^2+8x+12)
=(x+3)(x+5)(x+2)(x+6)

Answer 5

式子中^2表示平方 (x^2+4x+3)(x^2+12x+35)+15 =(x+1)(x+3)(x+5)(x+7)+15 =(x+1)(x+7)(x+5)(x+3)+15 =(x^2+8x+7)(x