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
<u>So...</u>
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
<u>So...</u>
A + B + 4 = 4
A + B = 0
<u>A = -B </u> ------------------------------------------------------------------------------------------------------------------- Part III - Solving for A and B with our two simultaneous equations
We know that<u> </u><u>A = -B</u><u /> 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
<u>A = 2</u>
Again, as we know that A = -B, and as we have found that A = 2, we can see:
A = -B
Therefore...
2 = -B
<u>B = -2 </u> So our final answer is <u>A = 2, B = -2</u><u />
Hopefully this answer is more useful than the last one, and isn't so confusing!
