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# 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) = kx^{2} - √2x + 1 (iv) p(x) = kx^{2} - 3x + k

**Solution:**

According to factor theorem, if x - 1 is a factor of p(x), then p(1) = 0

**(i)** p(x) = x^{2} + x + k

Since, x - 1 is a factor of the given polynomial p(x), thus p(1) = 0

⇒ p(1) = (1)^{2} + (1) + k

⇒ 0 = 2 + k

⇒ k = -2

**(ii)** p(x) = 2x^{2} + kx + √2

Since, x - 1 is a factor of the given polynomial p(x), thus p(1) = 0

⇒ p(1) = 2(1)^{2} + k(1) + √2

⇒ 0 = 2 + k + √2

⇒ k = -(2 + √2)

**(iii)** p(x) = kx^{2} - √2x + 1

Since, x - 1 is a factor of the given polynomial p(x), thus p(1) = 0

p(1) = k(1)^{2} - (√2 × 1) + 1

0 = k - √2 + 1

⇒ k = √2 - 1

**(iv)** p(x) = kx^{2} - 3x + k

Since, x - 1 is a factor of the given polynomial p(x), thus p(1) = 0

⇒ p(1) = k(1)^{2} - 3(1) + k

⇒ 0 = 2k - 3

⇒ k = 3/2

**☛ Check: **CBSE Class 9 NCERT Solutions Maths Chapter 2

**Video Solution:**

## 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

NCERT Solutions Class 9 Maths Chapter 2 Exercise 2.4 Question 3:

**Summary:**

The values of k, if (x - 1) is a factor of p(x) in each of the following p(x) = x^{2 }+ x + k, p(x) = 2x^{2} + kx + √2, p(x) = kx^{2} - √2x + 1, p(x) = kx^{2} - 3x + k are -2, -(2 + √2), √2 - 1, and 3/2 respectively.

**☛ Related Questions:**

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