# Polynomials

## Exercise 2.4 Part 3

Question 3: 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

Answer:To find zero of the given factor x – 1, we equate this with zero
Thus, x – 1 = 0
Or, x = 1
Since, x – 1 is a factor of p(x)
Therefore, p(1) = 0

Given, p(x) = x^2 + x + k
Or, p(1) = 1^2 + 1 + k
= 1 + 1 + k = 2 + k
Since, p(1) = 0
Hence, 2 + k = 0
Or, k = - 2

(ii) p(x) = 2x^2 + kx + sqrt2

Answer: To find zero of the given factor x – 1, we equate this with zero
Thus, x – 1 = 0
Or, x = 1
Since, x – 1 is a factor of p(x)
Therefore, p(1) = 0

Given, p(x) = 2x^2 + kx + sqrt2
Hence, p(1) = 2 xx 1^2 + k xx 1 + sqrt2
= 2 + k + sqrt2
Since, p(1) = 0
Hence, 2 + sqrt2 + k = 0
Or, k = - 2 - sqrt2 = - (2 + sqrt2)

(iii) p(x) = kx^2 - sqrt2x + 1

Answer:To find zero of the given factor x – 1, we equate this with zero
Thus, x – 1 = 0
Or, x = 1
Since, x – 1 is a factor of p(x)
Therefore, p(1) = 0

Given, p(x) = kx^2 - sqrt2x + 1
Hence, p(1) = k xx 1^2 - sqrt2 xx 1 + 1
= k - sqrt2 + 1
Since, p(1) = 0
Hence, k - sqrt2 + 1 = 0
Or, k = sqrt2 - 1

(iv) p(x) = kx^2 – 3x + k

Answer:To find zero of the given factor x – 1, we equate this with zero
Thus, x – 1 = 0
Or, x = 1
Since, x – 1 is a factor of p(x)
Therefore, p(1) = 0

Given, p(x) = kx^2 – 3x + k
Hence, p(1) = k xx 1^2 – 3 xx 1 + k
= k – 3 + k = 2k – 3
Since p(1) = 0
Hence, 2k – 3 = 0
Or, 2k = 3
Or, k = 3/2