Question

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 $
(ii) $ p(x) = 2{x^2} + kx + \sqrt 2 $
(iii) $ p(x) = k{x^2} - 2\sqrt x + 1 $
(iv) $ p(x) = k{x^2} - 3x + k $

Answer 1

Hint: : Remainder theorem states that, if a polynomial, $ p(x) $ is divided by a linear term $ (x - a) $ . Then the remainder of the division is $ p(a) $ . Factor theorem takes it further and states that, if $ (x - a) $ is the factor of a polynomial, $ p(x) $ . Then the remainder of the division is zero. i.e. $ p(a) = 0 $

Complete step-by-step answer:
It is given in the question that, $ (x - 1) $ is the factor of $ p(x) $ . That means, we can use the remainder theorem and the factor theorem and write,
 $ p(1) = 0 $ . . . (1)
(i) $ p(x) = {x^2} + x + k $
Put $ x = 1 $ , in the above equation,
 $ \Rightarrow p(1) = 1 + 1 + k $ . . . (2)
But from equation (1), we can write, $ p(1) = 0 $
Substituting this value in equation (2), we can write
 $ 0 = 1 + 1 + k $
By rearranging and simplifying it, we get
 $ k = - 2 $
(ii) $ p(x) = 2{x^2} + kx + \sqrt 2 $ . . . (3)
Put $ x = 1 $ , in the above equation,
 $ \Rightarrow p(1) = 2{(1)^2} + k(1) + \sqrt 2 $
But from equation (1), we can write, $ p(1) = 0 $
Substituting this value in equation (3), we can write
 $ 0 = 2{(1)^2} + k(1) + \sqrt 2 $
By rearranging and simplifying it, we get
 $ \therefore k = - 2 - \sqrt 2 $
(iii) $ p(x) = k{x^2} - \sqrt 2 x + 1 $ . . . (4)
Put $ x = 1 $ , in the above equation,
 $ \Rightarrow p(1) = k - \sqrt 2 + 1 $
But from equation (1), we can write, $ p(1) = 0 $
Substituting this value in equation (4), we can write
 $ 0 = k - \sqrt 2 + 1 $
By rearranging and simplifying it, we get
 $ k = \sqrt 2 - 1 $
(iv) $ p(x) = k{x^2} - 3x + k $
Put $ x = 1 $ , in the above equation,
 $ \Rightarrow p(1) = k(1) - 3(1) + k $
But from equation (1), we can write, $ p(1) = 0 $
Substituting this value in equation (4), we can write
 $ 0 = k(1) - 3(1) + k $
 $ \Rightarrow 0 = k - 3 + k $
By rearranging and simplifying it, we get
 $ 0 = 2k - 3 $
 $ \Rightarrow 2k = 3 $
 $ \Rightarrow k = \dfrac{3}{2} $

Note: We can also solve these questions by using the properties of root. Since, $ (x - 1) $ is a factor of $ p(x) $, $ x = 1 $ must be a root of the polynomial $ p(x) $ . That means, by using the property of roots, we can write that, $ p(1) = 0 $ .