# 3. Find the value of k, if :r - 1 is a factor of p(x) in each of the following cases: (i) p(:r) = x2 + % + k (iii) p(:r) = h2 - ./2% + 1 (u) p(x) = 2x2 + kx + ./2 (iv) p(x) = h2 - 3x + k.

Pawan Prajapati
3 years ago
(i) If (x - 1) is a factor of p(x). then p(l) = 0. Now, p(l) = 0 1 + 1 + k = 0 k = - 2. (u) If (x - 1) is a factor of p(x), then p(l) = O. Now, p(l) = 0 . = 2 + k + J2 = 0 = k =-(2 + J2 ). (ih) If (x - 1) is a factor of p(x), then p(l) = O. Now, p(l) = 0 => k ..... J2 + 1= 0 = k = J2 - 1. 34 (iv) If (x - 1) is a factor of p(x), then p(l) = 0. Now, p(l) = 0 4. Factorise: => k - 3 + k • O => k •i3
Ram Kushwah
110 Points
3 years ago
Dear student
The actual question is:
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)=kx²−2​x+1
(iv)  p(x)=kx²−3x+k

Solutions:

If x-1 is a factor of p(x)
Then as per factor theorem p(1)=0
Thus let us put x=1  and  p(1)=0 in all the given polynomials

(i) p(x) =x2+x+k
p(1) =1²+1+k=0
1+1+k=0
k=-2

(ii) p(x)=2x²+kx+2
p(1) =2*1²+k*1+2
2+k+2=0
k=-4

(iii) p(x)=kx²−2​x+1
p(1)=k*(1)²-2*1+1
k-2+1=0
k=2-1=1

(iv)  p(x)=kx²−3x+k
p(1)=k*(1)²-3*1+k=0
k-3+k=0
2k=3
k=3/2