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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

一般不认为含。要不就乱套了。我算的a是3.

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

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]

x^5-5x^4+ 8x^3-8x^2 +7x-3 =x^5-3x⁴-2x⁴+6x³+2x³-6x²-2x²+6x+x-3 =x⁴(x-3)-2x³(x-3)+2x²(x-3)-2x(x-3)+(x-3) =(x-3)(x⁴-2x³+2x²-2x+1) =(x-3)【(x⁴-2...

x^4+x^3+x^2+x+1 =x^2[x^2+x+1+1/x+1/x^2] =x^2[(x+1/x)^2+(x+1/x)-1] =x^2[(x+1/x)^2+(x+1/x)+1/4-5/4] =x^2[(x+1/x+1/2)^2-5/4] =x^2(x+1/x+1/2-√5/2)(x+1/x+1/2+√5/2) =(x^2+1+x/2-√5x/2)(x^2+1+x/2+√5x/2)

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-...

解:(x^3+x^2)+(x+1)=0 x^2(x+1)+(x+1)=0 (x+1)(x^2+1)=0 x^2+1=0 x^2=-1 x^2>=0 -1=0,不可能为负数, 或者说是x^2

x^3+3x^2+3x-26 =(x+1)^3-3^3 =(x+1-3)((x+1)^2+3(x+1)+3^2) =(x-2)(x^2+5x+13) x^4-4x+3 x^4-4x+3 =(x^4-x)-(3x-3) =x(x^3-1)-3(x-1) =x(x-1)(x^2+x+1)-3(x-1) =(x-1)(x^3+x^2+x-3) =(x-1)[(x^3-1)+(x^2-1)+(x-1)] =(x-1)[(x-1)(x^2+x+1)+(x-1)...

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