(x-1-√2i)(x-1+√2i)(x-2+√3i)(x-2-√3i)
问题描述:
(x-1-√2i)(x-1+√2i)(x-2+√3i)(x-2-√3i)
答
(x-1-i√2)(x-1+i√2)(x-2+i√3)(x-2-i√3)
=[(x-1)^2+2][(x-2)^2+3)]
=(x^2-2x+3)(x^2-4x+7)
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答
用平方差公式(a+b)(a-b)=a的平方-b的平方
所以
(x-1-√2i)(x-1+√2i)(x-2+√3i)(x-2-√3i)
=[(x-1)的平方-(√2i)的平方][(x-2)的平方-(√3i)的平方]
=(x^2-2x+3)(x^2-4x+7)
=x^4-6x^3+18x^2-26x+31