The Remainder Theorem: remainder when P(x) is divided by (x-1)(x+1)

ikegrd

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Jul 25, 2017
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Hello,

I've done a few remainder theorem problems in a workbook involving a single factor, i.e., "If polynomial P(x) is divided by x-r, its remainder is P(r)." In such a case, finding the remainder is easy. I need help applying "When polynomial P(x) is divided by D(x), it can be expressed as P(x) = Q(x)*D(x)+R(x)."

The problem at hand is:

A polynomial P(x) has a remainder of 2 when divided by (x-1) and a remainder of -2 when divided by (x+1). If P(x) is divided by (x-1)(x+1), then the remainder is
A. 2
B. x+2
C. 2x+1
D. 2x
E. 2x-1

The solution in my workbook states that the answer is D, where

P(x) = Q(x)(x-1)(x+1) + a(x-1) + 2
P(-1) = -2 = a(-1 -1) +2
a = 2

It then states that

The remainder is 2(x-1) + 2 = 2x when P(x) is divided by (x-1)(x+1)

Could someone help me understand what's going on here? Where is the "a" coming from in the solution? (I understand that the remainder must be first degree since the divisor is a second degree polynomial).

Much appreciated,
ikegrd
 
How do you determine the Remainder if you do NOT know that it has anything to do with P(r)?
 
Polynomial long division I suppose would give you the remainder as well..
 
Okay, so when a polynomial is divided by something (in this case (x-1)(x+1)), the quotient and possibly a remainder will result. The measure of divisibility of the divisor into the dividend is given by Q(x) + R. So in this case, P(x) / (x-1) = some Q(x) + 2, and P(x) / (x+1) = maybe some other Q(x) + -2.
 
Ah, so it just re-clicked that Q(x)*D(x) + R(x) is clearly equal to P(x). So this clearly gets me to P(x) = Q(x)*(x-1)(x+1) + R(x). It follows that that same P(x) = Q(x)*(x-1) + R(x), and P(1) = Q(x)*(x-1) + 2. P(-1) = Q(x)*(x+1) - 2.

Rearranging the equation, P(-1)/(x+1) - 2/(x+1) = Q(x)... I'm not sure if this is getting me anywhere but it feels like the right direction?
 
Ah, so it just re-clicked that Q(x)*D(x) + R(x) is clearly equal to P(x). So this clearly gets me to P(x) = Q(x)*(x-1)(x+1) + R(x). It follows that that same P(x) = Q(x)*(x-1) + R(x), and P(1) = Q(x)*(x-1) + 2. P(-1) = Q(x)*(x+1) - 2.

Rearranging the equation, P(-1)/(x+1) - 2/(x+1) = Q(x)... I'm not sure if this is getting me anywhere but it feels like the right direction?
Follow it; see where it goes.... ;)
 
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