Remainder Theorem Thinking Question

In summary, the remainder when a polynomial is divided by (x-1)(x+2) can be found by substituting x=1 and x=-2 into P(x)=(x-1)f(x)+2 and P(x)=(x+2)g(x)-19 to obtain two equations, and solving for a and b. In this case, the remainder is 7x-5.
  • #1
fakecop
42
0

Homework Statement


When a polynomial is divided by (x+2), the remainder is -19. When the same polynomial is divided by (x-1), the remainder is 2. Determine the remainder when the polynomial is divided by (x-1)(x+2)

Homework Equations


The Attempt at a Solution


had the polynomial been a real number, this problem would have been handled with modular mathematics (I think it's called the chinese remainder theorem?) But it isn't-it's polynomial division.

I began by stating that P(-2)=-19 and P(1)=2.

But then I got stuck-there is no way to finding any of the coefficients, because I only have the remainder for two values of x, and the degree of the polynomial isn't specified!

For example, if we assume the polynomial to be cubic. Then we have ax^3+bx^2+cx+d = R(x). I would then substitute the values for x and the remainders for R(x). But there are four unknowns-a, b, c, and d. I can't even solve this system by assuming that the polynomial is cubic, let alone not even knowing the degree.

Now I have come up with some polynomial expressions, such as x^3+4x-3 or 2x^3+x-1 which satisfy the criteria given in the question. Dividing by x^2+x-2, I found the remainder to be 7x-5. (Which is the answer in the book)

But how do I prove that for the general case? (Note that the degree of the polynomial isn't even specified!) Please help.
 
Last edited:
Physics news on Phys.org
  • #2
fakecop said:

Homework Statement


When a polynomial is divided by (x+2), the remainder is -19. When the same polynomial is divided by (x-1), the remainder is 2. Determine the remainder when the polynomial is divided by (x-1)(x+2)


Homework Equations





The Attempt at a Solution


had the polynomial been a real number, this problem would have been handled with modular mathematics (I think it's called the chinese remainder theorem?) But it isn't-it's polynomial division.

I began by stating that P(-2)=-19 and P(1)=2.

But then I got stuck-there is no way to finding any of the coefficients, because I only have the remainder for two values of x, and the degree of the polynomial isn't specified!

For example, if we assume the polynomial to be cubic. Then we have ax^3+bx^2+cx+d = R(x). I would then substitute the values for x and the remainders for R(x). But there are four unknowns-a, b, c, and d. I can't even solve this system by assuming that the polynomial is cubic, let alone not even knowing the degree.

Now I have come up with some polynomial expressions, such as x^3+4x-3 or 2x^3+x-1 which satisfy the criteria given in the question. Dividing by x^2+x-2, I found the remainder to be 7x-5. (Which is the answer in the book)

But how do I prove that for the general case? (Note that the degree of the polynomial isn't even specified!) Please help.

Call your polynomial P(x). The remainder when you divide P(x) by (x-1)(x+2) is going to be some linear polynomial ax+b. So you have P(x)=(x-1)f(x)+2, P(x)=(x+2)g(x)-19 and P(x)=(x-1)(x+2)h(x)+ax+b. Put x=1 and x=(-2) into those and see if you can get two equations to solve for a and b.
 
  • #3
wow I spent a week thinking about that question and you just helped me solve it in a few minutes.

So I get a+b=2 and -2a+b=-19.
solving, we get a=7 and b=-5. R(x)=7x-5.

So happy to see a simple solution, Thank you!
 
  • #4
fakecop said:
wow I spent a week thinking about that question and you just helped me solve it in a few minutes.

So I get a+b=2 and -2a+b=-19.
solving, we get a=7 and b=-5. R(x)=7x-5.

So happy to see a simple solution, Thank you!

You're welcome. Actually, coming up with sample polynomials that satisfied the premise and working out the remainder was pretty resourceful. That's a good way to attack a problem.
 

Related to Remainder Theorem Thinking Question

What is the Remainder Theorem?

The Remainder Theorem is a mathematical concept that helps us find the remainder when a polynomial is divided by a linear term. It states that if a polynomial f(x) is divided by (x-a), the remainder will be equal to f(a).

How is the Remainder Theorem used?

The Remainder Theorem is used to find the remainder when a polynomial is divided by a linear term. This can be useful in various mathematical applications, such as finding factors of polynomials, solving equations, and finding roots of polynomials.

What is the difference between the Remainder Theorem and the Factor Theorem?

The Remainder Theorem and the Factor Theorem are closely related but have some key differences. The Remainder Theorem helps us find the remainder when a polynomial is divided by a linear term, while the Factor Theorem helps us determine if a given number is a root of a polynomial. Additionally, the Factor Theorem is a special case of the Remainder Theorem, as the remainder will always be equal to zero when a given number is a root of a polynomial.

Can the Remainder Theorem be applied to any polynomial?

Yes, the Remainder Theorem can be applied to any polynomial, as long as the divisor is a linear term (of the form x-a). However, it is important to note that the remainder may not always be a whole number or a polynomial itself, as it can also be a fraction or a decimal.

What are some real-life applications of the Remainder Theorem?

The Remainder Theorem has various real-life applications, such as in coding theory, where it is used to generate error-correcting codes. It is also used in engineering, physics, and economics to model and solve various problems. In addition, the Remainder Theorem is commonly used in calculus to find derivatives and integrals of polynomials.

Similar threads

  • Precalculus Mathematics Homework Help
Replies
7
Views
778
  • Precalculus Mathematics Homework Help
Replies
4
Views
2K
  • Precalculus Mathematics Homework Help
Replies
3
Views
930
  • Precalculus Mathematics Homework Help
Replies
14
Views
1K
  • Precalculus Mathematics Homework Help
Replies
4
Views
892
  • Precalculus Mathematics Homework Help
Replies
15
Views
2K
  • Precalculus Mathematics Homework Help
Replies
33
Views
3K
  • Precalculus Mathematics Homework Help
Replies
7
Views
1K
  • Precalculus Mathematics Homework Help
Replies
7
Views
2K
  • Precalculus Mathematics Homework Help
Replies
9
Views
2K
Back
Top