Add maths (remainder theorem)

The polynomial x³ + ax² + bx - 3 leaves a remainder of 27 when divided by x-2 and a remainder of 3 when divided by x+1 . Calculate the remainder when the polynomial is divided by x-1... (the given answer is 3) please give a clear working, thanks in advance.. ∩_∩

You need to find a and b using the first information

when x = 2

8 + 4a +2b - 3 = 27
4a + 2b = 22.............[A]

when x = -1

-1 +a - b - 3 = 3
a-b = 7.....................[B]

multiply [B] by 2 and add to [A]
2a - 2b = 14
6a = 36
a = 6

from [B]
6-b = 7
b =- 1

the equation is

x^3 + 6x^2 - x - 3
If you put x = 1 into the found equation and put it equal to 0
f(x) = 1 + 6 - 1 - 3 = 3
= 3 too many

Let P(x) = x^3 + ax^2 + bx -3
When P (x) is divided by (x-2), the remainder is 27
==> P(2) = 8 + 4a + 2b - 3 = 27 (substituting x = 2 in the given polynomial)
==> 4a + 2b = 27 +3 -8 ==> 4a + 2b = 22
Dividing this equation by 2 , we get 2a + b = 11 .......... (1)

When P(x) is divided by (x+1), the remainder is 3
==> -1 + a - b -3 = 3 ==> a - b = 3 + 3 + 1 =7
==> a - b = 7 ............... (2)
Adding (1) and (2) gives 3a = 18 ==> a = 18/3 = 6
substituting in (1), 2*6 + b = 11 ==> b = 11 - 12 = -1
Therefore, the given polynomial is x^3 + 6x^2 - x - 3
When the polynomial is divided by (x - 1), the remainder is P(1)
==> P(1) = 1 + 6 -1 -3 = 7 - 4 = 3

The first statement means that for some polynomial P(x), x^3 + ax^2 + bx - 3 = P(x)(x-2) + 27. So if you plug in x = 2, you get:

27 = 8 + 4a + 2b - 3

Likewise for x=-1, 3 = -1 + a - b - 3.
Solve the system of equations to get a = 6, b = -1.
So the last equation (x = 1) is:

answer = 1 + a + b - 3
=> answer = 3.