Remainder Theorem Thinking Question

AI Thread Summary
When a polynomial is divided by (x+2) and (x-1), the remainders are -19 and 2, respectively. To find the remainder when divided by (x-1)(x+2), the polynomial can be expressed as a linear function ax + b. By substituting x = 1 and x = -2 into the equations derived from the remainders, two equations can be formed: a + b = 2 and -2a + b = -19. Solving these equations yields a = 7 and b = -5, resulting in the remainder R(x) = 7x - 5. This method effectively demonstrates how to handle polynomial division with unspecified degrees.
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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.
 
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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.
 
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!
 
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.
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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