Question

Find k so that 3x3 – 2kx2 + (k-1)x + 10 has (x+2) as a factor.​

Answer 1

Answer: k = -6/5

Step-by-step explanation:
(Given)
3x3 – 2kx2 + (k-1)x + 10 is divided by (x+2) then it should be equal to zero to get unknown variable K
Here (x+2)=0, then x=-2
The substitute x=-2 value in the polynomial equation.
Then we get
3x^3 - 2kx^2 + (k-1)x + 10=0
3(-2)^3 - 2k(-2)^2 + (k-1)(-2)+ 10 =0
3(-8) - 2k(4) -2k+2+10=0
-24 - 8k --2k+12=0
-10k-12=0
10k+12=0
K=((-6)/5)

Answer 2

Answer:

Step-by-step explanation:

Let p ( x ) = 3x³ - 2kx² +( k - 1 ) x + 10 = 0

If ( x + 2) is the factor of p ( x ),

then x + 2 = 0 ⇒x = - 2.

Now substituting " x = - 2 " in p ( x ), we get

p ( - 2 ) = 3 ( - 2 )³ - 2k ( - 2 )²+ ( k - 1 ) ( - 2 ) + 10 = 0

⇒3 ( - 8 ) - 2k ( 4 ) + ( - 2k + 2 ) + 10 = 0 (∵ - × + = - ; - × - = + )

⇒ - 24 - 8k - 2k + 2 + 10 = 0 (∵+ × - = - ; - × + = - )

⇒ - 10k - 24 + 12 = 0

⇒ - 10k - 12 = 0

⇒ 10k = - 12

⇒ k = - 12 / 10

⇒ k = - 6 / 5

∴ k = - 6/ 5 is the answer.