Q.3. Find the value of k, if x - 1 is a factor of p(x) in each of the following cases:

(i) p(x) = x2 + x + k

(ii) p(x) = 2x2 + kx +  2

(iii) p(x) = kx2 - √2x + 1

(iv) p(x) = kx2 - 3x + k

Solution:

If x - 1 is a factor of polynomial p(x), then p(1) should be 0.

(i) p(x) = x2 + x + k

⇒ (1)2 + 1 + k = 0

⇒ 2 + k = 0

⇒ k = - 2

Therefore, value of k is -2.

(ii) p(x) = 2x2 + kx +  2

⇒ 2(1)2 + k(1) +  = 0

⇒ 2 + k + 2 = 0

⇒ k = -2 - 2= -(2 + 2)

Therefore, value of k is -(2 + 2).

(iii) p(x) = kx2 - 2x + 1

k(1)2 - 2(1) + 1 = 0

⇒ k - 2 + 1 = 0

⇒ k = 2 - 1

Therefore, value of k is 2 - 1.

(iv) p(x) = kx2 - 3x + k

k(1)2 + 3(1) + k = 0

⇒ k - 3 + k = 0

⇒ 2k - 3 = 0

⇒ k = 32

Therefore, value of k is 32.