F4 Math一問~~~~~

A)
As the question mentioned, 2 is a root of the equation
Therefore
Put x=2 into the equation
(2)^2 - 5(2) + k(1-2) = 0
4 - 10 + k(-1) = 0
-6 - k = 0
k = -6

B)
Put k=-6 into the equation
x^2 - 5x + (-6)(1-x) = 0
x^2 - 5x - 6 + 6x = 0
x^2 + x - 6 = 0
(x+3)(x-2) = 0
Therefore another root is -3

a. Substitute x=2 into x^2 -5x + k(1-x) =0
2^2 - 5(2) + k(1-2)=0
4-10-k=0
k= - 6

b. x^2-5x-6(1-x)=0
x^2+x-6=0
(x+3)(x-2)=0
So the another root is -3.