Question

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​

Answer 1

Answer:

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Step-by-step explanation:

(i) p(x) = x2 + 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) = 2x2 + 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) = kx2 - √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) = kx2 - 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

Answer 2

Answer:

Answer:

hope it's helpful

Mark me brainliest

Step-by-step explanation:

(i) p(x) = x2 + 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) = 2x2 + 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) = kx2 - √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) = kx2 - 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