F(x)=x^3+ax^2+bx+3 where a and b are constants. bx given that when f (x) is divided by (x+2) the remainder is 7, (a) show that 2a-b=6 given also that when f (x) is divided by (x-1) the remainder is 4. (b) find the value of a and the value of b
i don't know how solve it :

Part I - First synthetic division

You need to use synthetic division to come up with an expression for a and b:

(x + 2) is a factor, and the remainder is 7, so we can draw a synthetic division table...

coefficients = 1 for X^3; A for X^2; B for X^1; and 3

-2  |        1          A            B                 3
     
                        -2        -2(A-2)          4(A-2)-2B
 
            1        (A-2)    -2(A-2)+B      4(A-2)-2B + 3  
                            
                                    Remainder = 7

So...

4(A-2)-2B + 3 = 7

4 * (A - 2) - 2B + 3 = 7

4A - 8 - 2B =  4

4A - 2B      = 12

2A - B = 6
Proved


Part II - Second Synthetic Division

We draw another synthetic division table, this time with (x - 1), so the number on the left hand side will be +1

1  |        1          A            B                 3
     
                        1         (A+1)         A+B+1
 
             1      (A+1)      A+B+1       A+B+4 
                            
                                    Remainder = 4

So... 

A + B + 4 = 4

A + B = 0

A = -B


Part III - Solving for A and B with our two simultaneous equations

We know that A = -B and we also know that 2A - B = 6

Since we know that A is equal to -B We can substitute in A for -B, to get:

2A - B = 6

Therefore...

2A + A = 6

3A = 6

A = 2

Again, as we know that A = -B, and as we have found that A = 2, we can see:

A = -B

Therefore...

2 = -B

B = -2

So our final answer is A = 2, B = -2

Hopefully this answer is more useful than the last one, and isn't so confusing!