求解一道因式分解
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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