Question

换元法 (x+1)(x+3)(x+5)(x+7)+15

Answer 1

(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)+15×7+15

=(x²+8x)²+22(x²+8x)+120

=(x²+8x+12)(x²+8x+10)

=(x+2)(x+6)((x²+8x+16)-6)

=(x+2)(x+6)(x+4+√6)(x+4-√6);

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