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.
Found 2 solutions by ankor@dixie-net.com, Edwin McCravy:
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! 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
Answer by Edwin McCravy(20060)  (Show Source):
You can put this solution on YOUR website! Find the quadratic
let the quadratic be Q(x) = Ax²+Bx+C
which has a remainder of -6 when divided by x-1,
By the remainder theorem Q(1) = A(1)²+B(1)+C = 6
A + B + C = 6
a remainder of -4 when divided by x-3,
By the remainder theorem Q(3) = A(3)²+B(3)+ C = 6
9A + 3B + C = 6
and no remainder when divided by x+1.
By the remainder theorem Q(-1) = A(-1)²+B(-1)+C = 0
A - B + C = 0
So solve this system of equations:
The substitute the values you get for A,B, and C in
Q(x) = Ax²+Bx+C
and that'll be the answer.
Edwin
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