x3 3x2 3x k has a factor of x 1, find k

Solution 1: x +3x -3x+k

x +x +2x +2x-5x-5+5+kx (x+1)+2(x+1)-5(x+1)+(k+5), x3+3x2-3x+k has a factor of x+1, k+5=0 , k=-5.

Solution 2: Let x3+3x2-3x+k=0 , x3+3x2-3x+k have a factor of x+1, x+1=0 , x=-1 , substitution, get:

1)³+3(-1)²-3(-1)+k=0 ， k=-5 。

The 1984 one was wrong, according to you.

It should be 3x (x-1) instead of x+1

I calculated that it was also k=-5

x^3+3x^2-3x+k

x^3+x^2+2x^2-3x+k

x 2 (x+1) + 2x 2-3x-5 + 5 + kx 2 (x+1) + (x+1) (2x-5) + (k+5) has a factor of x+1

Then k+5=0, k=-5

x=-1 is a solution of the polynomial to 0, which is directly substituted.

1+3+3+k=0

k=-5 doesn't have to be factored at all.

x 3 3x 2-3x+k has a cause of the Lu formula is x+1, then -1 is a root of x 3 silver limb 3x 2-3x+k=0, and the -1 generation x 3 socks are 3x 2-3x+k=0 to get k=-5

Reams x33x2

3x+k=(x+1)a, when x=-1, -1+3+3+k=0, the cluster is early.

Solve the cover Zheng closed k = -5

The answer is: -5

Let x 3 +3x 2 -3x+k=(x+1)a, when x=-1, -1+3+3+k=0, and get Wang which k=-5

Therefore, the answer is: Dongling Ze Na Peng-5

Is the formula wrong? The second item is good spring x finger trembling socks rent, the number of rents is 2, otherwise it can be combined.

One, two.

Since the polynomial has a factor of x+1, when x=-1, the value of that factor is 0, so the value of the whole polynomial is also 0

Substituting x=-1 to the polynomial:

1+3+3+k=0

k=-5

In 2x 3-x 2-5x+k, there is a factor of (x-2), which means that 2x 3-x 2-5x+k is divisible by (x-2), and there are at least two ways to find it:

1. Factoring method:

2x^3-x^2-5x+k

2x^3-4x^2)+(3x^2-6x)+(x+k)

2x^2(x-2)+3x(x-2)+(x+k)

The first two groups have a factor x-2, so x+k is also divisible by x-2, obviously k=-2

2. Division with remainder: divided = divided * business Qique + remainder, divisible remainder = 0

Therefore, substituting x=2 into the above equation = 0 is obtained, that is, the dividable formula 2x 3-x 2-5x+k=2*2 3-2 2-5*2+k=0, the solution is obtained; k=-2

You can also find the cofix by dividing the polynomial by the polynomial, which can also be found by the equivalence = 0. ,6,2x^3-x^2-5x+k=(x-2)(ax^2+bx+c)

ax^3+bx^2+cx-2ax^2-2bx-2c

ax^3+(b-2a)x^2+(c-2b)x-2c

a=2 b-2a=-1,b=3

c-2b=-5,c=1

k=-2ck=-2,2, because the original form has a factor, so the original form must be split into.

2x(3)-4x(2) =2x(2)(x-2)

3x(2)-6x =3x(x-2)

x +k =x-2

Therefore, k = -2,1, divide by (x-2) by (x-2) by (x-2) to get another cause of spine side combustion (2x 2+3x+1).

Then we can see that in k=-2,0,2x(3)-x(2)-5x+k, there is a factor of (x-2), then why is the value of k -2?

3) (2) is cubic and square, because the superscript cannot be typed.

Is the emptiness you want to express talking about blindness (x-1) (x-3)=k (x-3)? If not, then.

According to your paraphrase, k=x 2-1

If there is a factor of x+1, it can be divided into (x+1) ().

When x=-1, the value of the equation is zero.

Substituting -1+3+3+k=0

k=-5

x 3 3x 2-3x+k has a factor of x+1, then -1 is a root of x 3 3x 2-3x+k=0, and the -1 generation x 3 3x 2-3x+k=0 is solved to k=-5

There is also a way to make x+1=0, so x 3+3x 2-3x+k=0 and x=-1Finally, k=-5 is obtained

Is the formula wrong? The second x index is 2, otherwise it can be combined.

One, two.

Because a polynomial has a factor of x+1, when x=-1, the value of that factor is 0, so the value of the whole polynomial is also 0

Substituting x=-1 to the polynomial:

1+3+3+k=0

k=-5