Question

分解因式（1）（x+1)(x+3)(x+5)(x+7)+15

Answer 1

解：原式=[ (x+1) (x+7) ] [ (x+3) (x+5) ] +15
=(x^2 +8x +7) (x^2 +8x +15) +15.

令 t =x^2 +8x +11,
则 原式=(t -4) (t +4) +15
=t^2 -1
=(t+1) (t-1)
=(x^2 +8x +12) (x^2 +8x +10)
=(x+2) (x+6) (x^2 +8x +10).

= = = = = = = = =
1. 换元法。
因为 x^2 +8x +7 和 x^2 +8x +15 的平均数为 x^2 +8x +11,
所以 令 t =x^2 +8x +11.

2. x^2 +8x +10 在有理数范围内不能分解。
因为 10=1*10=2*5, 而两组都不行.

Answer 2

解：因为：
(x+3)(x+5)=x^2+8x+15
(x+1)(x+7)=x^2+8x+7
设(x+3)(x+5)＝y
则：(x+1)(x+7)＝y-8

原式化为 y(y-8)+15
＝y^2-8y+15
＝(y-3)(y-5)
＝(x^2+8x+12)(x^2+8x+10)
＝(x+2)(x+6)(x^2+8x+10)

Answer 3

(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
=(x^2+8x+7)[(x^2+8x+7)+8]+15
=(x^2+8x+7)^2+8(x^2+8x+7)+15
=[(x^2+8x+7)+3][(x^2+8x+7)+5]
=(x^2+8x+10)(x^2+8x+12)
=(x^2+8x+10)(x+2)(x+6)

Answer 4

(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
设y=x^2+8x+11
原式
=(y-4)(y+4)+15
=y^2-16+15
=y^2-1
=(y+1)(y-1)
=(x^2+8x+12)(x^2+8x+10)
=(x+2)(x+6)(x^2+8x+10)

Answer 5

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