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`



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