DIvide x3+x2-14x-24 by x+2Solution x2-x-12/)x-2)x3+x2-14x-24x3+2x/-2x2-14x-x2-2x/-12x-24-12x-24/0In this example we see that x+2 is a factor of x3+x2-14x-24,as the remainder is zero.Thus x3+x2-14x-24÷（x+2）=x2-x-12 with zero remainderso x3+x2-14x

来源：学生作业帮助网 编辑：作业帮 时间：2024/04/19 15:23:22

DIvide x3+x2-14x-24 by x+2Solution x2-x-12/)x-2)x3+x2-14x-24x3+2x/-2x2-14x-x2-2x/-12x-24-12x-24/0In this example we see that x+2 is a factor of x3+x2-14x-24,as the remainder is zero.Thus x3+x2-14x-24÷（x+2）=x2-x-12 with zero remainderso x3+x2-14x
DIvide x3+x2-14x-24 by x+2
Solution x2-x-12/)x-2)x3+x2-14x-24
x3+2x/-2x2-14x
-x2-2x/-12x-24
-12x-24/0
In this example we see that x+2 is a factor of x3+x2-14x-24,as the remainder is zero.
Thus x3+x2-14x-24÷（x+2）=x2-x-12 with zero remainder
so x3+x2-14x-12/x+2=x2-x-12

DIvide x3+x2-14x-24 by x+2Solution x2-x-12/)x-2)x3+x2-14x-24x3+2x/-2x2-14x-x2-2x/-12x-24-12x-24/0In this example we see that x+2 is a factor of x3+x2-14x-24,as the remainder is zero.Thus x3+x2-14x-24÷（x+2）=x2-x-12 with zero remainderso x3+x2-14x
（x³+x³-14x-24）÷（x+2)=x²-x-12.
由x³+x²-14x-24
=x³-x³-12x+2x²-2x-24
=x（x²-x-12）+2（x²-x-12）
=（x²-x-12）（x+2).

是考SAT吗？

DIvide x3+x2-14x-24 by x+2Solution x2-x-12/)x-2)x3+x2-14x-24x3+2x/-2x2-14x-x2-2x/-12x-24-12x-24/0In this example we see that x+2 is a factor of x3+x2-14x-24,as the remainder is zero.Thus x3+x2-14x-24÷（x+2）=x2-x-12 with zero remainderso x3+x2-14x 因式分解x3+5x2+2x-24 x3+9x2+26x+24x3+9x2+26x+24怎么写啊? 因式分解（1+x+x2+x3)2-x3 分解因式：-27x3+8 ,(x2-5x+2)(x2-5x+4)-24 ,x5+x4+x3+x2+x1+1 因式分解x3-x2-x-1 x3 -x2- x- 2 因式分解 shuxue (16 21:24:57)已知f(x)=x+x3,x1.x2.x3∈R且x1+x2,x2+x3,x1+x3均为正数,试判断f(x)+f(x2)+f(x3)的符号 计算(x3+x2+x+2)/(x2+1) 化简x3-x2/x2+x-2 lim_x→无穷(x3+2x2+x/2x3-x2) 因式分解：x4+7x3+14x2+7x+1 x4+7x3+14x2+7x+1 x3-9x2+26x-24怎么解,具体点. 解方程 x3-24x2+144x=121 练习一1．分解因式：(2)x10+x5-2；(4)(x5+x4+x3+x2+x+1)2-x5．2．分解因式：(1)x3+3x2-4；(2)x4-11x2y2+y2；(3)x3+9x2+26x+24；(4)x4-12x+323．3．分解因式：(1)(2x2-3x+1)2-22x2+33x-1；(2)x4+7x3+14x2+7x+1；(3)(x+y)3+2xy(1-x-y)-1；( (3X3+5X2+X)/(7X3+5X-4) x3+x+2 x3-4x2-4x-5 因式分解