Polynomial Question //////////

a)By remainder theorem ,f(k) = (k - m) (k - n) (k - 4) - 4 = - 2
==>
(k - m) (k - n) (k - 4) = 2

b)k - 4 is an integer must be a factor of 2 , therefore k - 4 = - 2 , - 1 , 1 or 2
k = 2 , 3 , 5 or 6Note that k - m > k - n > k - 4 since m < n < 4.When k = 2 ,
(2 - m) (2 - n) (2 - 4) = 2
(2 - m) (2 - n) (- 2) = 2
2 - m = 1 and 2 - n = - 1 for (2 - m) > (2 - n) > (- 2)
m = 1 and n = 3When k = 3 ,
(3 - m) (3 - n) (3 - 4) = 2
(3 - m) (3 - n) (- 1) = 2
No solutions for (3 - m) > (3 - n) > (- 1)When k = 5 or 6 ,
(5 - m) (5 - n) (5 - 4) = 2 or
(6 - m) (6 - n) (6 - 4) = 2
No solutions for (5 - m) > (5 - n) > (5 - 4) or (6 - m) > (6 - n) > (6 - 4)Hence k = 2 , m = 1 , n = 3

c)x² - 3x + 2 = (x - 1) (x - 2)[f(x)]²
= [ (x - 1) (x - 3) (x - 4) - 4 ]²By remainder theorem ,
When [f(x)]² is divided by x - 1 , the remainder is [f(1)]² = [0 - 4]² = 16
When [f(x)]² is divided by x - 2 , the remainder is [f(2)]² = [(1)(-1)(-2) - 4]² = 4Let [f(x)]² = Q(x) * (x - 1)(x - 2) + p(x - 1) + 16 ,then [f(2)]² = 0 + p(2 - 1) + 16 = 4 ,
p + 16 = 4
p = - 12When [f(x)]² is divided by (x - 1)(x - 2) , the remainder is p(x - 1) + 16
= - 12(x - 1) + 16
= - 12x + 28