用十字相乘法解:(x2+3x-3)(x2+3x+4)-8

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用十字相乘法解:(x2+3x-3)(x2+3x+4)-8

用十字相乘法解:(x2+3x-3)(x2+3x+4)-8
用十字相乘法解:(x2+3x-3)(x2+3x+4)-8

用十字相乘法解:(x2+3x-3)(x2+3x+4)-8
设x2+3x=t
(x2+3x-3)(x2+3x+4)-8
=(t-3)(t+4)-8
=t^2+t-12-8
=t^2+t-20
=(t+5)(t-4)
=(x2+3x+5)(x2+3x-4)
=(x2+3x+5)(x+4)(x-1)

(x2+3x-3)(x2+3x+4)-8
=(x2+3x)²-7﹙x2+3x﹚-8
=(x2+3x+8)×(x2+3x+1)

That"s all,thank you.