The polynomial \(\p(x)\) is of degree 9 and \(\p(x)-1\) is exactly divisible by \((x-1)^5\).
Find the value of \(\p(1)\).
Show that \(\p'(x)\) is exactly divisible by \((x-1)^4\).
Given also that \(\p(x)+1\) is exactly divisible by \((x+1)^5\), find \(\p(x)\).
Solution: \(p(x) = q(x)(x-1)^5 + 1\) where \(q(x)\) has degree \(4\).
\(p(1) = q(1)(1-1)^5 + 1 = 1\).
\(p'(x) = q'(x)(x-1)^5 + 5(x-1)^4q(x) + 0 = (x-1)^4((x-1)q'(x) + 5q(x))\) so \(p'(x)\) is divisible by \((x-1)^4\)
\(p(x)+1\) divisible by \((x+1)^5\) means that \(p(-1) = -1\) and \(p'(x)\) is divisible by \((x+1)^4\). Since \(p'(x)\) is degree \(8\) it must be \(c(x+1)^4(x-1)^4 = c(x^2 - 1)^4\).
Expanding and integrating, we get \(p(x) = c(\frac{1}{9}x^9 -\frac{4}{7}x^7 + \frac{6}{5}x^5 - \frac{4}{3}x^3 + x) + d\).
When \(x = 1\) we get \(c \frac{128}{315} + d = 1\) and when \(x = -1\) we get \(-c \frac{128}{315} + d = -1\) so \(2d = 0 \Rightarrow d = 0, c = \frac{315}{128}\) and
\[ p(x) =\frac{315}{128} \l \frac{1}{9}x^9 -\frac{4}{7}x^7 + \frac{6}{5}x^5 - \frac{4}{3}x^3 + x\r \]