2x三次方+x²-5x+2 因式分解?
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2x^3+x^2-5x+2
=2x^3+x^2-x -4x+2
=x(2x^2+x-1)-2(2x-1)
=x(x+1)(2x-1)-2(2x-1)
=(2x-1)(x^2+x-2)
=(2x-1)(x-1)(x+2),4,2x³+x²-5x+2=﹙2x³-2x²﹚+﹙3x²-5x+2﹚
=2x²﹙x-1﹚+﹙x-1﹚﹙3x-2﹚
=﹙x-1﹚﹙2x²+3x-2﹚
=﹙x-1﹚﹙x+2﹚﹙2x-1﹚,1,2x³+x²-5x+2
=(2x²+x-6)+2/x+x
=(x+3)(x-2)+2/x+x
,1,观察2x^3+x^2-5x+2,我们会发现当x=1时代数式的系数和恰好为0,所以2x^3+x^2-5x+2的一个因子为(x-1),即
2x^3+x^2-5x+2=(x-1)(2x^2+3x-2)=(x-1)(2x-1)(x+2),1,2x³+x²-5x+2
=(2x³-2x²)+(3x²-5x+2)
=2x²(x-1)+(3x-2)(x-1)
=(x-1)(2x²+3x-2)
=(x-1)(2x-1)(x+2),0,2x三次方+x²-5x+2 因式分解
=2x^3+x^2-x -4x+2
=x(2x^2+x-1)-2(2x-1)
=x(x+1)(2x-1)-2(2x-1)
=(2x-1)(x^2+x-2)
=(2x-1)(x-1)(x+2),4,2x³+x²-5x+2=﹙2x³-2x²﹚+﹙3x²-5x+2﹚
=2x²﹙x-1﹚+﹙x-1﹚﹙3x-2﹚
=﹙x-1﹚﹙2x²+3x-2﹚
=﹙x-1﹚﹙x+2﹚﹙2x-1﹚,1,2x³+x²-5x+2
=(2x²+x-6)+2/x+x
=(x+3)(x-2)+2/x+x
,1,观察2x^3+x^2-5x+2,我们会发现当x=1时代数式的系数和恰好为0,所以2x^3+x^2-5x+2的一个因子为(x-1),即
2x^3+x^2-5x+2=(x-1)(2x^2+3x-2)=(x-1)(2x-1)(x+2),1,2x³+x²-5x+2
=(2x³-2x²)+(3x²-5x+2)
=2x²(x-1)+(3x-2)(x-1)
=(x-1)(2x²+3x-2)
=(x-1)(2x-1)(x+2),0,2x三次方+x²-5x+2 因式分解
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