Polynomial Division which leaves remainders

[potato124]

I do not need anyone to give me any answer(because who wants to do anyone's work). I just simply want some links that anyone could provide some knowledge needed to solve these types of questions. Thanks in advance.

 

[Steven G]

P(x) leaves a remainder -x+5 when divided by x^2-2x-8 implies that P(x) is a cubic equation.
P(x) leaves a remainder x when divided by x^2-1also implies that P(x) is a cubic equation.

Therefore P(x) = ax^3 + bx^2 + cx + d

Do the divisions and using the given remainders solve for a, b, c and d

 

[HallsofIvy]

First, since dividing P by quadratic polynomials leaves first degree remainders, P itself must be cubic: \(\displaystyle P(x)= ax^3+ bx^2+ cx+ d\)
\(\displaystyle \frac{ax^3+ bx^2+ cx+ d}{x^2- 2x- 8}= Q(x)- x+ 5\) so [

 

[blamocur]

I might be missing something here, but would not any polynomial (i.e., not just a cubic one) leave a 1st degree remainder when divided by a 2nd degree polynomial ?

 

[Steven G]

Not necessarily, the remainder could be a 1st degree equation or a constant.

Consider (x^4 +2x^3 -7x^2 + 4)/(x^2) = x^2 + 2x + 4/(x^2). The remainder 4 is not 1st degree.

Sometimes the remainder will be 0!

It is true, that the remainder will be at most 1st degree.

 

[blamocur]

Not necessarily, the remainder could be a 1st degree equation or a constant.

Consider (x^4 +2x^3 -7x^2 + 4)/(x^2) = x^2 + 2x + 4/(x^2). The remainder 4 is not 1st degree.

Sometimes the remainder will be 0!

It is true, that the remainder will be at most 1st degree.

Agree. But the original polynomial can be any degree above 0, not just cubic.

 

[blamocur]

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I do not need anyone to give me any answer(because who wants to do anyone's work). I just simply want some links that anyone could provide some knowledge needed to solve these types of questions. Thanks in advance.

The only useful link I can think of is to the definition of the polynomial division with remainder (e.g., https://en.wikipedia.org/wiki/Polynomial#Division). Then some not very complicated tricks can be used to get the answer.