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I'm aware that in remainder theorem you take the divisor and make it equal to zero. For the result of that just plug in x into whatever the polynomial dividend is.

But its different when the divisor is also a polynomial since it has multiple roots:

$$p(x) = \frac{2x^6 - x^4 - 4x^2 - 8}{x^2 + 4x - 3}$$

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2 Answers 2

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$x^2+4x-3$ into $2x^6-x^4-4x^2-8$.

First, look at leading terms. $x^2$ into $2x^6$. That's $2x^4$. Multiply divisor ($x^2+4x-3$) by $2x^4$ (getting $2x^6+8x^5-6x^4$) and subtract from dividend ($2x^6-x^4-4x^2-8$), getting $-8x^5+5x^4-4x^2-8$. Leading terms again: $x^2$ into $-8x^5$ is $-8x^3$. Multiply divisor by $-8x^3$, subtract from $-8x^5+5x^4-4x^2-8$. Write the whole thing out the way you would write out $123$ into $456789$ and you'll see we're just doing division but with polynomials instead of numbers. Keep going until what's left of the dividend has degree less than that of the divisor, that is, less than 2. At that point what's left is the remainder, and the quotient is $2x^4-8x^3+$ whatever other terms you have worked out along the way.

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When you divide by a polynomial of degree greater than one things get tricky. Especially if the polynomial you are dividing by has double roots, but let's ignore that case because its quite tricky. Suppose to you are dividing $P(x)$ by $Q(x)$ which has roots $r_1,r_2,...,r_n$. Compute $P(r_1),P(r_2),...,P_(r_n)$. Then apply Lagrange Interpolation on $(r_1,P(r_1)),...,(r_n,P(r_n))$ to get the remainder. Do you see why this works?

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Wouldn't it be easier to just perform synthetic division? –  EuYu Nov 15 '12 at 4:01
Of course that would be much easier, however it appears mwong wants a method similar to the remainder theorem so I gave him the most natural way possible to extend to higher degree. –  dinoboy Nov 15 '12 at 4:08
I would mark this answer as correct if I knew what the Lagrange Interpolation was. Would it possible to provide the answer in long or synthetic division so it would be something that I can understand? @EuYu –  mwong Nov 15 '12 at 4:12
Alternatively, I could tell you what Lagrange interpolation is.... –  Gerry Myerson Nov 15 '12 at 4:29

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