Let \(f(x)\) be a \(5^\text{th}\) degree polynomial such that \(f(x) + 1 \) is divisible by \((x-1)^3\), and \(f(x)-1\) is divisible by \((x+1)^3\). Find \(f(x) \).

We have \( f(x) + 1 = (x-1)^3 g(x) \) and \( f(x) -1 = (x+1)^3h(x) \). Thus, we want to find degree 2 polynomials such that \( g(x) (x-1)^3 - h(x) (x+1)^3 =2 \).

We know that \((x-1)\) is a factor of \(f(x)+1.\)Using factor and remainder theorem,we get
\(f(1)+1=0\)
\(f(1)=-1\)
We know that \((x+1)\) is a factor of \(f(x)-1.\)Using factor and remainder theorem,we get
\(f(-1)-1=0\)
\(f(-1)=1\)
while seeing the two equations we can guess that the leading co-efficient is 1 and that the polynomial is \(\boxed {-x^5}.\)

Unfortunately \( -x^5 + 1 \) is not a multiple of \( (x-1)^3\).

You had a good start, applying remainder factor theorem. You applied it to just \( (x-1)\) and so you got 1 condition. You should also apply it to \( (x-1)^2\) and \( (x-1)^3\) to get more conditions. Ditto for \( (x+1)^3\). This gives you 6 conditions, and we have 6 unknowns, so we should be able to solve that.

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TopNewestWe have \( f(x) + 1 = (x-1)^3 g(x) \) and \( f(x) -1 = (x+1)^3h(x) \). Thus, we want to find degree 2 polynomials such that \( g(x) (x-1)^3 - h(x) (x+1)^3 =2 \).

This is just bezout's identity applied to polynomials.

\( (x+1)^3 = 1 \times (x-1)^3 + (6x^2 + 6) \)

\( (x-1)^3 = (\frac{1}{6}x-\frac{1}{2}) \times (6x^2 + 6) +(2x+2) \)

\( (6x^2 +6) = (3x-3)(2x+2) + 12 \)

So we apply the backwards step of euclidean algorithm to obtain

\( 2 = 1 \times (x^2 + 1) - (\frac{x-1}{2}) \times (2x+2) \\ = 1 \times (x^2 + 1) - ( \frac{x-1}{2} ) \times [ (x-1)^3 - (x-3)(x^2+1) ] \\ = ( \frac{ x^2 -4x+5} { 2} ) \times (x^2 + 1) - ( \frac{x-1}{2} ) \times (x-1)^3 \\ = ( \frac{ x^2 -4x+5} { 2} ) \times \frac{1}{6} [ (x+1)^3 - (x-1)^3] - ( \frac{x-1}{2} ) \times (x-1)^3 \\

= \frac{ x^2-4x+5}{12} \times (x+1)^3 - \frac{x^2+2x-1}{12} \times (x-1)^3 \\ \)

Thus, \( f(x) = \frac{ x^2-4x+5}{12} \times (x+1)^3 -1 \).

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Totallly forgot about euclidian's algorithm.

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We know that \((x-1)\) is a factor of \(f(x)+1.\)Using factor and remainder theorem,we get

\(f(1)+1=0\)

\(f(1)=-1\)

We know that \((x+1)\) is a factor of \(f(x)-1.\)Using factor and remainder theorem,we get

\(f(-1)-1=0\)

\(f(-1)=1\)

while seeing the two equations we can guess that the leading co-efficient is 1 and that the polynomial is \(\boxed {-x^5}.\)

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Unfortunately \( -x^5 + 1 \) is not a multiple of \( (x-1)^3\).

You had a good start, applying remainder factor theorem. You applied it to just \( (x-1)\) and so you got 1 condition. You should also apply it to \( (x-1)^2\) and \( (x-1)^3\) to get more conditions. Ditto for \( (x+1)^3\). This gives you 6 conditions, and we have 6 unknowns, so we should be able to solve that.

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But my answer is \(-x^5\) and not \(-x^5+1.\)

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What have you tried? What are your thoughts?

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I'm trying it out.I have no idea right now.

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