(i) p(x) = x² + x + k Putting x - 1 = 0, we get, x = 1 Using remainder theorem, when p(x) = x² + x + k is divided by x - 1, remainder is given by p(1) = (1)² + (1) + k = 2 + k Since x - 1 is a factor of p(x), hence remainder p(1) = 0 ⇒ 2 + k = 0 ⇒ k = -2
(i) p(x) = x² + x + k
Putting x – 1 = 0, we get, x = 1
Using remainder theorem, when p(x) = x² + x + k is divided by x – 1, remainder is given by p(1)
= (1)² + (1) + k
= 2 + k
Since x – 1 is a factor of p(x), hence remainder p(1) = 0
⇒ 2 + k = 0
⇒ k = -2
(ii) p(x) = 2x² + kx + √2 Putting x - 1 = 0, we get, x = 1 Using remainder theorem, when p(x) = 2x² + kx + √2 is divided by x - 1, remainder is given by p(1) = 2(1)² + k(1) + √2 = 2 + k + √2 Since x - 1 is a fctor of p(x), hence remainder p(1) = 0 ⇒ 2 + k + √2 = 0 ⇒ k = -2 - √2
(ii) p(x) = 2x² + kx + √2
Putting x – 1 = 0, we get, x = 1
Using remainder theorem, when p(x) = 2x² + kx + √2 is divided by x – 1, remainder is given by p(1)
= 2(1)² + k(1) + √2
= 2 + k + √2
Since x – 1 is a fctor of p(x), hence remainder p(1) = 0
⇒ 2 + k + √2 = 0
⇒ k = -2 – √2
(iii) p(x) = kx² – √2 x + 1 Putting x - 1 = 0, we get, x = 1 Using remainder theorem, when p(x) = kx² – √2 x + 1 is divided by x - 1, remainder is given by p(1) = k(1)² – √2(1) + 1 = k – √2 + 1 Since x - 1 is a factor of p(x), hence remainder p(1) = 0 ⇒ k - √2 + 1 = 0 ⇒ k = √2 - 1
(iii) p(x) = kx² – √2 x + 1
Putting x – 1 = 0, we get, x = 1
Using remainder theorem, when p(x) = kx² – √2 x + 1 is divided by x – 1, remainder is given by p(1)
= k(1)² – √2(1) + 1
= k – √2 + 1
Since x – 1 is a factor of p(x), hence remainder p(1) = 0
⇒ k – √2 + 1 = 0
⇒ k = √2 – 1
(iv) p(x) = kx² - 3x + k Putting x - 1 = 0, we get, x = 1 Using remainder theorem, when p(x) = kx² - 3x + k is divided by x - 1, remainder is given by p(1) = k(1)² - 3(1) + k = 2k - 3 Since x - 1 is a factor p(x), hence remainder p(1) = 0 ⇒ 2k - 3 = 0 ⇒ k = 3/2
(iv) p(x) = kx² – 3x + k
Putting x – 1 = 0, we get, x = 1
Using remainder theorem, when p(x) = kx² – 3x + k is divided by x – 1, remainder is given by p(1)
= k(1)² – 3(1) + k
= 2k – 3
Since x – 1 is a factor p(x), hence remainder p(1) = 0
⇒ 2k – 3 = 0
⇒ k = 3/2
Find the value of k, if x – 1 is a factor of p(x) in the following cases: p(x) = x² + x + k
(i) p(x) = x² + x + k Putting x - 1 = 0, we get, x = 1 Using remainder theorem, when p(x) = x² + x + k is divided by x - 1, remainder is given by p(1) = (1)² + (1) + k = 2 + k Since x - 1 is a factor of p(x), hence remainder p(1) = 0 ⇒ 2 + k = 0 ⇒ k = -2
(i) p(x) = x² + x + k
See lessPutting x – 1 = 0, we get, x = 1
Using remainder theorem, when p(x) = x² + x + k is divided by x – 1, remainder is given by p(1)
= (1)² + (1) + k
= 2 + k
Since x – 1 is a factor of p(x), hence remainder p(1) = 0
⇒ 2 + k = 0
⇒ k = -2
Find the value of k, if x – 1 is a factor of p(x) in the following cases: p(x) = 2x² + kx + √2
(ii) p(x) = 2x² + kx + √2 Putting x - 1 = 0, we get, x = 1 Using remainder theorem, when p(x) = 2x² + kx + √2 is divided by x - 1, remainder is given by p(1) = 2(1)² + k(1) + √2 = 2 + k + √2 Since x - 1 is a fctor of p(x), hence remainder p(1) = 0 ⇒ 2 + k + √2 = 0 ⇒ k = -2 - √2
(ii) p(x) = 2x² + kx + √2
See lessPutting x – 1 = 0, we get, x = 1
Using remainder theorem, when p(x) = 2x² + kx + √2 is divided by x – 1, remainder is given by p(1)
= 2(1)² + k(1) + √2
= 2 + k + √2
Since x – 1 is a fctor of p(x), hence remainder p(1) = 0
⇒ 2 + k + √2 = 0
⇒ k = -2 – √2
Find the value of k, if x – 1 is a factor of p(x) in the following cases: p(x) = kx² – √2 x + 1
(iii) p(x) = kx² – √2 x + 1 Putting x - 1 = 0, we get, x = 1 Using remainder theorem, when p(x) = kx² – √2 x + 1 is divided by x - 1, remainder is given by p(1) = k(1)² – √2(1) + 1 = k – √2 + 1 Since x - 1 is a factor of p(x), hence remainder p(1) = 0 ⇒ k - √2 + 1 = 0 ⇒ k = √2 - 1
(iii) p(x) = kx² – √2 x + 1
See lessPutting x – 1 = 0, we get, x = 1
Using remainder theorem, when p(x) = kx² – √2 x + 1 is divided by x – 1, remainder is given by p(1)
= k(1)² – √2(1) + 1
= k – √2 + 1
Since x – 1 is a factor of p(x), hence remainder p(1) = 0
⇒ k – √2 + 1 = 0
⇒ k = √2 – 1
Find the value of k, if x – 1 is a factor of p(x) in the following cases: p(x) = kx² – 3x + k
(iv) p(x) = kx² - 3x + k Putting x - 1 = 0, we get, x = 1 Using remainder theorem, when p(x) = kx² - 3x + k is divided by x - 1, remainder is given by p(1) = k(1)² - 3(1) + k = 2k - 3 Since x - 1 is a factor p(x), hence remainder p(1) = 0 ⇒ 2k - 3 = 0 ⇒ k = 3/2
(iv) p(x) = kx² – 3x + k
See lessPutting x – 1 = 0, we get, x = 1
Using remainder theorem, when p(x) = kx² – 3x + k is divided by x – 1, remainder is given by p(1)
= k(1)² – 3(1) + k
= 2k – 3
Since x – 1 is a factor p(x), hence remainder p(1) = 0
⇒ 2k – 3 = 0
⇒ k = 3/2