Question

Find the value of k if x – 3 is a factor of k²x square -kx2 + 3kx - k.


need the ans.....plzz help me to solve

​

Answer 1

Answer:

$\large{\underline{\boxed{\sf k = \dfrac{1}{9}}}}$

Step-by-step explanation:

Given that ;

(x - 3) is a factor of the given polynomial (k²x² - kx² + 3kx - k).

⇒ x - 3 = 0

⇒ x = 4

Substitute the value of x in the polynomial and compare it with zero.

⇒ k² (3)² - k(3)² + 3 * k * 3 - k = 0

⇒ 9k² - 9k + 9k - k = 0

⇒ 9k² - k = 0

⇒ k(9k - 1) = 0

⇒ 9k - 1 = 0

⇒ 9k = 1

⇒ k = $\sf \dfrac{1}{9}$

_______________________

Hence, the value of k is $\bf \dfrac{1}{9}$

Answer 2

Answer:

$k = \frac{4}{3}$

Step-by-step explanation:

given :-

x - 3 is a factor

so it  divides the given expression exactly leaving remainder zero

x - 3 = 0

x = 3

Given quadratic equation :-

$k^{2} x^{2} - kx^{2} +3kx-k$

substute " x" value

[$k^{2} x^{2} - kx^{2} + 3kx-k \\\\k^{2}3^{2} -k3^{2} +3k(3) - k = 0\\9k^{2} - 9k+ 9k - k =0\\\\9k^{2} - k =0\\\\k ( 9k - 1 ) = 0$

now we have two cases

case - 1

k = 0

case -2

9k - 1 = 0

9k = 1

$k = \frac{1}{9}$

===========================================

thanks for the question.