Question

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

(i) p(x) = x ² + x + k 

(ii) p(x) = 2x ² + kx +√2 

(iii) p(x) = kx² – \(\sqrt{2x}\) + 1 

(iv) p(x) = kx² – 3x + k

Answer 1

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

(i) p(x) = x² + x + k p(1) = 0 

⇒(1) 2 + 1 + k = 0 

⇒(2) + k = 0 k = −2. 

Therefore, the value of k is -2.

(ii) p(x) = 2x² + kx + √2 = p(1) = 0 

⇒ 2 (1)² + k (1) + √2 = 0 

⇒ 2 + k + √2 = 0 

⇒ k = -2 - √2 = -(2 + √2). 

Therefore, the value of k is -(2 + √2).

(iii) p(x) = kx² - √2x + 1 = p(1) = 0 

⇒ k (1)² - √2(1) + 1 = 0 

⇒ k - √2 + 1 = 0 

⇒ k = √2 - 1.

(iv) p(x) = kx² - 3x + k 

⇒ p(1) = 0 

⇒ k(1)² - 3(1) + k = 0 

⇒ k = -3 + k = 0 

⇒ 2k - 3 = 0 

⇒ k = \(\frac{3}{2}\)

Therefore, the value of k is \(\frac{3}{2}\).