Question

plz solve this fast.
ans is k= -3
after dividing i m getting remainder as 2k^2+21x+7kx+8k+6.

Answer 1

Given f(x) = 2x^4 + x^3 - 14x^2 + 5x + 6.

Given g(x) = x^2 + 2x + k.

Now,

Divide f(x) by g(x).



                           2x^2  -  3x - (8 + 2k)
                        ----------------------------------------------------------------
x^2  +  2x +  k) 2x^4  + x^3  -  14x^2 +   5x       +      6

                         2x^4  -  4x^3 +  2kx^2

                          ----------------------------------------------------------

                                    - 3x^3  - (14+ 2k)x^2 + 5x

                                    -3x^2 - 6x^2

                           -----------------------------------------------------------

                                             -(8 + 2k)x^2 +  (5 + 3k)x  +  6

                                             -(8 + 2k)x^2  - (16 + 4k)x  - (8k + 2k^2)

                           -------------------------------------------------------------------------

                                                                     (21 + 7k)x +  (2k^2 + 8k + 6)



Therefore Remainder = (21 + 7k) + (2k^2 + 8k + 6) = 0.

Now,

21 + 7k = 0

7k = -21

k = -3.

Therefore the value of k = -3.


We know that:

Dividend = Divisor * Quotient + remainder

0 = (x^2 + 2x + k) * (2x^2 - 3x - 8 - 2k) + 0

0 = (x^2 + 2x - 3) * (2x^2 - 3x - 8 - 2(-3)) 

0 =  (x^2 + 2x - 3) * (2x^2 - 3x - 2)

0 = (x^2 + 3x - x - 3) * (2x^2 - 4x + x - 2)

0 = (x(x + 3) - 1(x + 3) * (2x(x - 2) + 1(x - 2))

0 = (x -1 )(x + 3)(2x + 1)(x - 2)

x = 1, -3, -1/2, 2.


Therefore the other zeroes of the polynomial are 1,-3,-1/2,2.


Hope this helps!

Answer 2

Hey friend, Harish here

Here is your answer.

To Find:

The value of K.

Solution:

When we divide the two equations the remainder you are getting is ( 21 + 7k)x + (2k²+ 8k + 6) which is correct. And the quotient is 2x² - 3x - (8+2k) 

Given that, x² + 2x + k  is a factor. Then remainder must be Zero .

Then, (21 + 7k)x = 0  -(i) and (2k²+ 8k + 6) = 0. -(ii)

From (i) We get : 

⇒ 21 + 7k = 0 

⇒ 7k = -21

⇒ $k = \frac{-21}{7} = -3$

Now the factors of the expression are:

x² + 2x - 3  and  2x² - 3x -2. 

Because Dividend = Divisor × Quotient + remainder (0) .

Then, 

⇒ (x² + 2x - 3)(2x² - 3x -2)

⇒ (x-1)(x+3) × (x-2)(2x+1)

This Equation Becomes Zeroes during the following cases:

i) (x-1) = 0   ⇒ x = 1

ii) (x+3)= 0 ⇒ x = -3

iii) (x-2) = 0 ⇒ x = 2

iv) (2x+1) = 0 ⇒ x = -1/2

$\boxed{\bold{Therefore \ the \ zeroes\ are\ 1, -3,2,\frac{-1}{2} }}$
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I Hope my answer is helpful to you.