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初二数学因式分解:若x^3+5x^2+7x+A有一因式x+1,...

您好,(x+1)(x^2+bx+c) =x^3+bx^2+cx+x^2+bx+c =x^3+(b+1)x^2+(b+c)x+c=X^3+5x^2+7x+a 系数对应相等 b+1=5 b+c=7 c=a 解出 b=4,c=a=3 所以a=3 X^3+5x^2+7x+3 =(x+1)(x^2+4x+3) =(x+1)^2(x+3)

答案是

x^4+7x^3-12x^2+7x+| =(x^2+1)^2+7x(x^2+1)-14x^2 以上式子可以分解,但比较麻烦 ,此题的-12x^2可能是12x^2 x^4+7x^3+12x^2+7x+| =(x^2+1)^2+7x(x^2+1)+10x^2 =[(x^2+1)+5x][(x^2+1)+2x] =(x^2+5x+1)(x+1)^2

(x^2-5x+4)(x^2+9x+18)+108 =x^4+4x^3-23x^2-54x+180 =x^3(x-3)+7x^2(x-3)-2x(x-3)-60(x-3) =(x-3)(x^3+7x^2-2x-60) =(x-3)[x^2(x+5)+2x(x+5)-12(x+5)] =(x-3)(x+5)(x^2+2x-12)

设(x+1)^2=y,那么原式整理后 =[(x+1)^2+1]^2+4[3(x+1)^2+x][2(x+1)^2+(x-1)] =(y+1)^2+4(3y+x)[2y+(x-1)] =y^2+2y+1+4[6y^2+(5x-3)y+x(x-1)] =25y^2+(20x-10)y+4x^2-4x+1 =(5y)^2+10(2x-1)y+(2x-1)^2 =[5y+(2x-1)]^2 =[5(x+1)^2+(2x-1)]^2 =(5x^...

=x的3次方(x+1)+(x+1)(x-1) =(x+1)(x的3次方+x-1)

解法一: x³-3x²+(a+2)x-2a =x³-3x²+2x+ax-2a =x(x²-3x+2)+a(x-2) =x(x-1)(x-2)+a(x-2) =(x-2)[x(x-1)+a] =(x-2)(x²-x+a) 解法二: x³-3x²+(a+2)x-2a =x³-2x²-x²+(a+2)x-2a =x²(x-...

这个多项式在有理数范围内已经不能再分解了 可以先用二次方程求根公式求出该多项式的两个零点,x1=(5+√13)/2,x2=(5-√13)/2, 故x^2-5x+3=[x-(5+√13)/2][x-(5-√13)/2]

不知道你是怎么解的,这儿有一种解法: x^4-7x2+1=(x2+1)2-9x2=(x2+1+3x)(x2+1-3x) 然后自己自己处理哈

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