Remainder Theorem with 2 unknowns.

In summary, when rx^3 + gx^2 +4x + 1 is divided by x-1 and x+3, the remainders are 12 and -20 respectively. The values of r and g can be found by solving the equations 12=r+g+9 and -20=r+g using elimination, which yields the solution r=12 and g=-9.
  • #1
naffle
2
0

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!
 
Physics news on Phys.org
  • #2
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:
Code:
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.
Code:
 27r+27g= 81
-27r+ 9g=-13
------------------
  0r+36g= 68
 
Last edited:
  • #3
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!
 

Related to Remainder Theorem with 2 unknowns.

What is the Remainder Theorem with 2 unknowns?

The Remainder Theorem with 2 unknowns is a mathematical concept that helps in finding the remainder when a polynomial is divided by a binomial with 2 unknowns. It is used to simplify complex polynomial expressions and solve equations involving two variables.

How does the Remainder Theorem with 2 unknowns work?

The Remainder Theorem states that when a polynomial P(x, y) is divided by a binomial (x-a, y-b), the remainder is equal to P(a, b). This means that if we substitute the values of a and b in the polynomial, we will get the remainder. This theorem is based on the Division Algorithm.

What are the applications of the Remainder Theorem with 2 unknowns?

The Remainder Theorem with 2 unknowns is used in various fields of science and engineering, such as physics, chemistry, and computer science. It is used to solve systems of equations, find the roots of polynomial equations, and perform polynomial division to simplify complex expressions.

Can the Remainder Theorem with 2 unknowns be used for polynomials with more than 2 variables?

Yes, the Remainder Theorem can be extended to polynomials with any number of variables. In fact, the Remainder Theorem with 2 unknowns is a special case of the more general Multivariable Remainder Theorem, which can be applied to polynomials with any number of variables.

Are there any limitations to the Remainder Theorem with 2 unknowns?

Yes, the Remainder Theorem with 2 unknowns can only be applied to binomials with 2 unknowns. It cannot be used for polynomials with more than 2 variables or for binomials with only 1 unknown. Additionally, the theorem only works for exact division, meaning that the remainder will be 0 if the binomial is a factor of the polynomial.

Similar threads

  • Precalculus Mathematics Homework Help
Replies
3
Views
906
  • Calculus and Beyond Homework Help
Replies
6
Views
1K
  • Calculus and Beyond Homework Help
Replies
3
Views
921
  • Calculus and Beyond Homework Help
Replies
1
Views
1K
  • Calculus and Beyond Homework Help
Replies
2
Views
2K
  • Precalculus Mathematics Homework Help
Replies
14
Views
1K
  • Calculus and Beyond Homework Help
Replies
24
Views
3K
  • Calculus and Beyond Homework Help
Replies
1
Views
2K
  • Calculus and Beyond Homework Help
Replies
12
Views
1K
  • Calculus and Beyond Homework Help
Replies
6
Views
2K
Back
Top