F4 A-MATHS 急......

Let
f(r) = r^3 + (2-α)r^2 + (β-2α+1)r - (α+1)
f(2) = 0 and f'(2) = 0
then
8 + 4(2-α) + 2(β-2α+1) - (α+1) = 0
12 + 4(2-α) + (β-2α+1) = 0
Hence,
α = 25/3
β = 29

Here is a method WITHOUT DIFFERENTIATION:

r^3 + (2-α)r^2 + (β-2α+1)r - (α+1) = 0

r =2 is a double root.
(r-2)^2 is a factor of r^3 + (2-α)r^2 + (β-2α+1)r - (α+1)

Therefore,
r^3 + (2-α)r^2 + (β-2α+1)r - (α+1) = (r-2)2(r-c)
r^3 + (2-α)r^2 + (β-2α+1)r - (α+1) = r^3 - (c+4)r^2 + (4c+4)r - 4c

Compare the r^2 term: 2-α = -(c+4) …… (1)
Compare the r term: β-2α+1 = 4c+4 …… (2)
Compare the constant term: -(α+1) = -4c …… (3)

(1):
2-α = -c-4
c =α-6 …… (4)
4c = 4α-24…… (5)

(3)
4c =α+1 ….. (6)

(5)=(6)
4α-24 = α+1
3α = 25
α = 25/3

Put α=25/3 into (4)
c = (25/3)-6
c = 7/3

Put α=25/3 and c=7/3 into (2):
β-2(25/3)+1 = 4(7/3)+4
β-47/3 = 40/3
β = 29

Ans: α = 25/3, β = 29

Cubic equations should be something in pure maths rather than amaths...