（y+1）的4次方+（y+3)的4次方—272的因式分解
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解决时间 2021-04-14 23:30
- 提问者网友：爱了却不能说
- 2021-04-14 00:45
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最佳答案
- 五星知识达人网友：何以畏孤独
- 2021-04-14 02:22
（y+1)^4+(y+3)^4-272
=[(y+1)^2-(y+3)^2]^2+2(y+1)^2 (y+3)^2-272
=[(y+1+x+3)（y+1-y-3)]^2+2[(y+1)(y+3)]^2-272
=[2(y+2)*(-2)]^2+2[(y+2-1)(y+2+1)]^2-272
=16(y+2)^2+2[(y+2)^4-2(y+2)^2+1]-272
=2(y+4)^4+12(y+2)^2-270
=2[(y+4)^4+6(y+2)^2-135]
=2[(y+2)^2-9][(y+2)^2+15)
=2(y+2+3)(y+2-3)(y^2+4y+19)
=2(y+5)(y-1)(y^2+4y+19)
=[(y+1)^2-(y+3)^2]^2+2(y+1)^2 (y+3)^2-272
=[(y+1+x+3)（y+1-y-3)]^2+2[(y+1)(y+3)]^2-272
=[2(y+2)*(-2)]^2+2[(y+2-1)(y+2+1)]^2-272
=16(y+2)^2+2[(y+2)^4-2(y+2)^2+1]-272
=2(y+4)^4+12(y+2)^2-270
=2[(y+4)^4+6(y+2)^2-135]
=2[(y+2)^2-9][(y+2)^2+15)
=2(y+2+3)(y+2-3)(y^2+4y+19)
=2(y+5)(y-1)(y^2+4y+19)
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