因式分解:(2x+5)(x^2-9)(2x-7)-91 要用换元法!

来源:学生作业帮助网 编辑:作业帮 时间:2024/08/01 14:14:10
因式分解:(2x+5)(x^2-9)(2x-7)-91 要用换元法!
xJ@_D]xI7y$h#)<-BQh\r f'b>I\ޗE-Ǧ!nj)I]6`egT?/_r~V}\m#vE֢T)rRa`a^˧tcfZ`#lN3]:+ߦI).w6D2H# :_%s

因式分解:(2x+5)(x^2-9)(2x-7)-91 要用换元法!
因式分解:(2x+5)(x^2-9)(2x-7)-91 要用换元法!

因式分解:(2x+5)(x^2-9)(2x-7)-91 要用换元法!
(2x+5)(x^2-9)(2x-7)-91
=(2x+5)(x-3)(x+3)(2x-7)-91
=[(2x+5)(x-3)][(x+3)(2x-7)]-91
=(2x^2-x-15)(2x^2-x-21)-91
设2x^2-x-15=y
则原式=y(y-6)-91
=y^2-6y-91
=(y-13)(y+7)
所以原式=(x^2-x-15-13)(x^2-x-15+7)
=(x^2-x-28)(x^2-x-8)

(2x-7)(2x+5)(x^2-9)-91
=(2x-7)(x+3)(2x+5)(x-3)-91
=(2x^2-x-21)( 2x^2-x-15)-91
设 2x^2-x=a
原式=(a-21)(a-15)-91
=a²-36a+315-91
=a²-36a+224
=(a-28)(a-8)
原式等于(2x^2-x-28)(2x^2-x-8)