
因式分解:(x +1)的4次方+(x+3)的4次方—272
1个回答
2014-02-21
展开全部
(x+1)^4+(x+3)^4-272
=[(x+1)^2-(x+3)^2]^2+2(x+1)^2 (x+3)^2-272
=[(x+1+x+3)(x+1-x-3)]^2+2[(x+1)(x+3)]^2-272
=[2(x+2)*(-2)]^2+2[(x+2-1)(x+2+1)]^2-272
=16(x+2)^2+2[(x+2)^4-2(x+2)^2+1]-272
=2(x+4)^4+12(x+2)^2-270
=2[(x+4)^4+6(x+2)^2-135]
=2[(x+2)^2-9][(x+2)^2+15)
=2(x+2+3)(x+2-3)(x^2+4x+19)
=2(x+5)(x-1)(x^2+4x+19)
不知道对不对,没什么把握
=[(x+1)^2-(x+3)^2]^2+2(x+1)^2 (x+3)^2-272
=[(x+1+x+3)(x+1-x-3)]^2+2[(x+1)(x+3)]^2-272
=[2(x+2)*(-2)]^2+2[(x+2-1)(x+2+1)]^2-272
=16(x+2)^2+2[(x+2)^4-2(x+2)^2+1]-272
=2(x+4)^4+12(x+2)^2-270
=2[(x+4)^4+6(x+2)^2-135]
=2[(x+2)^2-9][(x+2)^2+15)
=2(x+2+3)(x+2-3)(x^2+4x+19)
=2(x+5)(x-1)(x^2+4x+19)
不知道对不对,没什么把握
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