Question 3. Find the Value of K, If X – 1 is a Factor of P(em Exercise 2.4 Chapter 2 Polynomial Maths Class 9 NCERT

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) = x² + x + k
(ii) p(x) = 2x² + kx + √2
(iii) p(x) = kx² – 2x + 1
(iv) p(x) = kx² – 3x + k

Solution: (i) p(x) = x² + x + k
Apply remainder theorem
=>x - 1 =0
=> x =  1
According to remainder theorem p(1) = 0 we get
Plug x = 1 we get  
=> k(1)²  + 1+ 1 =0
=>k  +1 + 1 =0
=>  k  + 2 = 0
=> k = - 2 
Answer value of k = -2

(ii) p(x) = 2x² + kx + √2
Apply remainder theorem
=>x - 1 =0
=> x =  1
According to remainder theorem p(1) = 0 we get
Plug x = 1 we get  
p(1) = 2(1)² + k(1) + √2
p(1) =2 +  k  + √2
0 = 2 + √2 + k
-2 - √2 = k
- (2 + √2)  = k  
Answer is k = - (2 + √2)

(iii) p(x) = kx2 – √2x + 1
Apply remainder theorem
=>x - 1 =0
=> x =  1
According to remainder theorem p(1) = 0 we get
Plug x = 1 we get  
p(1) = k(1)² – √2(1)+ 1
P(1) = K - √2  + 1
0 = K - √2  + 1
√2  -1 = K
Answer k= √2 -1

(iv) p(x) = kx² – 3x + k
Apply remainder theorem
=>x - 1 =0
=> x =  1
According to remainder theorem p(1) = 0 we get
Plug x = 1 we get 
P(1) = k(1)² -3(1) + k
0= k – 3 + k
0 = 2k – 3
3 = 2k
3/2 = k

Answer k = 3/2