Question 1147584
Find the quadratic which has a remainder of -6 when divided by x-1, a remainder of -4 when divided by x-3, and no remainder when divided by x+1.
:
Use the form ax^2 + bx + c, we assume a=1
Synthetic division; divisor is (x-1) remainder is -6
....__________
 1 | 1 + b + c
...........1+(b+1)
....------------
...... 1+(b+1)- 6; the equation from this
(b+1)+c = -6
b + c = -7
:
divisor is (x-3) remainder is -4
....__________
3 | 1 + b + c
...........3+(3b+9)
....------------
......1+(b+3) - 4; the equation from this
3b+9 + c = -4
3b + c = -4 - 9
3b + c = -13
:
Use elimination with these two equations to find b
3b + c = -13
 b + c = -7
----------------subtraction eliminates c find b
2b + 0 = -6
b = -3
find c when b=-3
-3 + c = -7
c = -7 + 3
c = -4
:
The equation: y = x^2 - 3x - 4
:
:
you can confirm this:
perform long division using divisors (x-1) and (x-3) to give remainders of  -6 and -4, respectively