Please explain how to answer this polynomials question

The polynomial x^3+8x^2+ax+b has x−1 as a factor.
When the polynomial is divided by (x−2), there is a remainder of 40. Find the values of a and b.

So this is what I did
1^3=b*1^2+a+b
9+a+b=0
a=-b-9

f(2)=40
(2)^3+8(3)^2+2a+b=40
80+2a+b=40

2a+b=-40
2(-b-9)+b=-40
-2b-18+b=-40
-b=-22

which is obviously completely wrong since my answers are no where near the actual answers

the answers are supposed to be a = 9, b = -18

please show in steps and explain

thank you

Because (x-1) is a factor of the original polynomial, when you use long division (or synthetic, whichever you prefer), the remainder is zero. Using long division the "remainder" expression you get from
x³ + 8x² + ax + b
------------------------
x - 2

is b + a + 9 = 0, or as you have above,
(I) a + b = -9.

Doing the same with (x-2), you get a remainder expression of b + 2a + 40, which you set equal to 40:
(II) 2a + b + 40 = 40
or, equivalently,
2a + b = 0

Solve the system for either a or b. I'll work with a:
a + b = -9
2a + b = 0

-a - b = 9
2a + b = 0
========
a = 9

a + b = -9
9 + b = -9
b = -18

We know that plugging in 2 for x makes the whole thing equate to 40

(2)^3 + 8x^2 + 2a + b = 40 right?

Simplifying a bit:

8 + 32 + 2a + b = 40
2a+b = 0

or, if you like
b = -2a

Next I used synthetic division in dividing the polynomial by (x-1)

Because x-1 is a factor then a zero is the remainder, if you will.

Examining the synthetic division setup it becomes obvious that

9+a+b had to equal zero.

So I had a system of equations:

9+a+b=0

b=-2a

Subbing -2a for "b" in the first equation yields:

9+a-2a=0
a=9

and so on