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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)

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

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-...

解法一: x³-6x²+15x-10 =x³-x²-5x²+5x+10x-10 =x²(x-1)-5x(x-1)+10(x-1) =(x-1)(x²-5x+10) 解法二: x³-6x²+15x-10 =x³-1-6x²+15x-9 =(x-1)(x²+x+1)-3(2x²-5x+3) =(x-1)(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]

(1)15a3+10a2=5a2(3a+2);(2)m(a-3)+2(3-a)=(a-3)(m-2);(3)x2+7x+10=(x+2)(x+5).故答案为:(1)5a2(3a+2),(2)(a-3)(m-2),(3)(x+2)(x+5).

一般不认为含。要不就乱套了。我算的a是3.

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^...

∵有一因式x+1 ∴方程x³+5x²+7x+a=0有根x=-1 ∴-1+5-7+a=0 a=3 ∴x*x*x+5x*x+7x+a =x³+5x²+7x+3 =x³+x²+4x²+7x+3 =x²(x+1)+(x+1)(4x+3) =(x+1)(x²+4x+3) =(x+1)(x+1)(x+3)

(x-1)(x+3)(x-2)(x+4)这样相乘可以看出来了哇? 把(x^2+2x)看做是一个整体 =(x^2+2x-3)(x^2+2x-8)+6 =(x^2+2x)^2-11(x^2+2x)+30 =(x^2+2x-5)(x^2+2x-6)

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