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求解一道因式分解

更新时间2018-07-16 09:49:02

因式分解题:5x²+x-2

解答:原式=5x²+x-2

               =5(x²+0.2x-0.4)

               =5(x²+0.2x+0.1²-0.1²-0.4)

               =5[(x+0.1)²-0.41]

               =5[(x+0.1)²-0.1√41]·[(x+0.1)²+0.1√41]。

 

5x²+x-2=5(x²+0.2x-0.4)=5(x²+0.2x+0.1²-0.1²-0.4)

=5[(x+0.1)²-0.41]=5[(x+0.1)²-0.1√41]·[(x+0.1)²+0.1√41]

回答完毕保证正确

5x²+x-2

=(1/20)(100x²+20x-40)

=(1/20){[(10x)²+2(10x)+1]-41}

=(1/20)[(10x+1)²-(√41)²]

=(1/20)(10x+1+√41)(10x+1-√41)

令5x²+x-2=0

求根公式解得

x=(-1±√41)/10


5x²+x-2

=5[x+(1+√41)/10][x+(1-√41)/10]


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