Question

一元二次整式因式分解具体公式公式

Answer 1

用配方法,亲自把公式推导出来,印象必然更深

ax" + bx + c

= a[ x" + ( b/a )x + ( c/a ) ]

= a[ x" + 2( b/2a )x + ( b/2a )" - ( b"/ 4a" ) + ( 4ac/4a" ) ]

= a{ [ x + ( b/ 2a ) ]" - [ ( b" - 4ac ) /4a" ] }

这样平方差,分解因式

= a{ [ x + ( b/ 2a ) ]" - [ √( b" - 4ac ) /2a ]" }

= a{ [ x + ( b/ 2a ) ] + [ √( b" - 4ac ) /2a ] }{ [ x + ( b/ 2a ) ] - [ √( b" - 4ac ) /2a ] }

= a{ x + [ b + √( b" - 4ac ) ] / 2a }{ x + [ b - √( b" - 4ac ) ] / 2a }

或者

= a{ x - [ -b + √( b" - 4ac ) ] / 2a }{ x - [ -b - √( b" - 4ac ) ] / 2a }

3x" + 4x - 1

= 3{ x + [ 4 + √( 16 + 12 ) ] / 6 }{ x + [ 4 - √( 16 + 12 ) ] / 6 }

= 3{ x + ( 4 + 2√7 ) / 6 }{ x + ( 4 - 2√7 ) / 6 }

= 3{ x + ( 2 + √7 ) / 3 }{ x + ( 2 - √7 ) / 3 }

= ( 1/3 )( 3x + 2 + √7 )( 3x + 2 - √7 )

x" - 5x + 3

= { x - [ 5 + √( 25 - 12 ) ] / 2 }{ x - [ 5 - √( 25 - 12 ) ] / 2 }

= { x - ( 5 + √13 ) / 2 }{ x - ( 5 - √13 ) / 2 }

= ( 1/4 )( 2x - 5 + √13 )( 2x - 5 - √13 )