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 $\frac{3}{2}$.