Remainder Theorem with 2 unknowns.

[naffle]

Homework Statement

When rx^3 + gx^2 +4x + 1 is divided by x-1, the remainder is 12. When it is divided by x+3, the remainder is -20. Find the values of r and g.

Homework Equations

The Attempt at a Solution

r=f(1)
=r(1)^3 + g(1)^2 + 4(1) +5
=r + g +9
r=12
r+g+9=12
r+g= 3

r=f(-3)
=r(-3)^3 + g(-3)^2 + 4(-3) +5
=-27r +9g -7
r=-20
-27r +9g -7=-20
-27 +9g=-13

Then I was just going to use elimination to find r and g.
-27r-27g=-81 (equation 1 multiplied by -27)
-27r+9g=-13
-------------
-36g=-68
g= 1.89

But that answer is wrong.
Is one of my equations wrong or am I going about it incorrectly?
I was able to find the unknown in a equation similar to this, so I'm not sure why I'm having such trouble with this one.
Thanks for any help!

 

[NateTG]

When rx^3 + gx^2 +4x + 1 is...
...
r=f(1)
=r(1)^3 + g(1)^2 + 4(1) +5
...

r=f(-3)
=r(-3)^3 + g(-3)^2 + 4(-3) +5

Where did the 5 come from?

NB:
It's probably a bad idea to use 'r' as your remainder as well as one of the unknowns.

r=f(1)
=r(1)^3 + g(1)^2 + 4(1) +5
=r + g +9
r=12
r+g+9=12
r+g= 3

Might be written better as:

12=f(1)
12=(1)^3r+(1)^2g+(1)4+5
12=1r+1g+4+5
12=r+g+9
 3=r+g

Also, this bit is correct:

-27r-27g=-81 (equation 1 multiplied by -27)
-27r+9g=-13
-------------
-36g=-68
g= 1.89

But IMHO this is much more legible.

 27r+27g= 81
-27r+ 9g=-13
------------------
  0r+36g= 68

 

[naffle]

Oh wow, you're completely right.
I think I just wrote the equation wrong in my notes, but when I posted this I read the equation from the textbook. Hahah, well, thanks!