Question

Find the values of 'k', if x-3 is a factor of k 2 x 3 -kx ​2 +3kx-k.

Answer 1

Answer:

K = 1/27

Step-by-step explanation:

Let p(x) = k²x³ - kx² + 3kx - k

Given x - 3 is a factor of k²x³ - kx² + 3kx - k

Then x - 3 = 0

Since, x = 3 is a zero of the p(x), then p(3) = 0

Thus, substituting in the equation -

= k²(3)³ - k(3)² + 3k(3) - k = 0

= 27k² - 9k + 9k - k = 0

= 27k² - k = 0

=k(27k - 1) = 0

Case 1, k = 0

Case 2, 27k - 1 = 0

= k = 1/27

Therefore, the required value of k is either 1/27 or 0.

Answer 2

Answer:

$Values \: of \: k \: are \: 0 , \frac{1}{27}$

Step-by-step explanation:

Let p(x)=k²x³-kx²+3kx-k,

If (x-3) is a factor of p(x) then p(3)=0 /* By factor theorem

Now,

k²(3)³-k(3)²+3k(3)-k=0

=> 27k²-9k+9k-k=0

=> 27k²-k=0

=> k(27k-1)=0

=> k = 0 Or 27k-1 = 0

=> k = 0 Or 27k = 1

$\implies k=0 \:Or \: k = \frac{1}{27}$

Therefore,

$Values \: of \: k \: are \: 0 , \frac{1}{27}$

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