x的平方减3x减10因式分得
X\u7684\u5e73\u65b9\u51cf3X\u51cf10\u56e0\u5f0f\u5206\u89e3\u5f97\uff08x\uff0b2\uff09\u00d7\uff08x-5\uff09
\u89e3\uff1a\u539f\u5f0f=x^4-5x^2+6-3x^2+10
=x^4-8x^2+16
=(x^2-4)^2
=(x-2)^2(x+2)^2
\u671b\u91c7\u7eb3
法1：构造法
x^2-3x-10
=x^2-5x+2x-10
=x(x-5)+2(x-5)
=(x+2)(x-5)
法2：十字相乘法
x^2-3x-10
x 2
x -5交叉相乘相加得-5x+2x=3x满足中间项
所以
x^2-3x-10
=(x+2)(x-5)
x^2-3x-10=(x-5)(x+2)
解：x^2-3x-10=(x+2)(x-5)=0
x=-2或x=5
绛旓細x^2-3x-10 =(x+2)(x-5)
绛旓細x^2-3x-10=(x+2)(x-5)x 2 x -5
绛旓細(x-5)(x+2)鈮0鎵浠鈮5鎴-2
绛旓細鍥炵瓟锛氳В:x^2+3x-10 =(x+5)(x-2)
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绛旓細鍘熷紡=x²-4x+x-4 =x(x-4)+(x-4)=(x-4)(x+1)
