Question

for what value of k is y3+ky+2k-2 exactly divisible by (y+1)​

Answer 1

Answer:

The value of k is 3

Step-by-step explanation:

p(y)=y³+ky+2k-2

p(-1)=(-1)³+k×(-1)+2k-2

=-1-k+2k-2

=-3+k

-3+k=0

k=3

so the value of k is 3

Answer 2

The correct answer is 3.

Given: The equation = $y^{3}+ky+2k-2$.

Factor which is exactly divisible = y + 1.

To Find: The value of k.

Solution:

If (y + 1) is exactly divisible by the equation.

y + 1 = 0

y = -1

By putting y = -1 the equation should be equal to 0.

$y^{3}+ky+2k-2$ = 0

$(-1)^{3}+(-1)k+2k-2=0$

-1 -k +2k -2 = 0

k = 3

Hence, the value of k is 3.

#SPJ2