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I have been given that a polynomial $f(x)$ with real coefficients is divisible by $(x^2+1)$, and that when $f'(x)$ is divided by $(x^2+1)$, we get a remainder of $(x+1)$. I need to prove that $2f(x)+(x-1)(x^2+1)$ is divisible by $(x^2+1)^2$.

What I did was to express $f(x)$ as $f(x)=(x^2+1)g(x)$, for some polynomial $g$ (1),and $f'(x)=((x^2+1))q(x) + (x+1)$, for some polynomial $q$ (2). From (1) I deduced another expression of $f'(x)$, (3): $$f'(x)=(x^2+1)g'(x) + 2x(g(x))$$ I then expressed the given expression as: $$2f(x)+(x-1)(x^2+1) = 2(x^2+1)g(x) + (x-1)(x^2+1)$$ (4) To find an expression for $g(x)$, I then equated (2) and (3) above, to get $$ 2g(x) = (x^2+1)(q(x)-g'(x) + x+1)(1/x)$$

Letting $(q(x)-g'(x) + x+1)(1/x) = h(x)$, I simplified to: $$2g(x)= (x^2+1)h(x)$$ then here I stopped, realizing that $h(x)$ is maybe not even a polynomial. Any alternative method or advancement of this one will be very appreciated!

2good4this
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2 Answers2

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Mark's answer is nicer than this, but it is possible to continue along the lines you started:

Since $f'(x)=(x^2+1)g'(x)+2x g(x)$, $2x g(x)$ has the same remainder $(x+1)$ when divided by $x^2+1$ as does $f'(x)$. That is, $$ 2xg(x)=h(x)(x^2+1)+x+1 $$ for some polynomial $h$. By setting $x=0$, we can see that $h(0)=-1$; thus $h(x)=xk(x)-1$ for some other polynomial $k$. So \begin{align} 2xg(x)&=(xk(x)-1)(x^2+1)+x+1\\ &=xk(x)(x^2+1)-x^2+x \end{align} Dividing by $x$ gives $$ 2g(x)=k(x)-(x-1) $$ which upon multiplying by $(x^2+1)$ yields the desired relation.

Micah
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You can use the rule for a linear factor that $a$ is a double-root, and $(x-a)^2$ a factor, of $p(x)=0$ if and only if $p(a)=p'(a)=0$.

Use $a=i$ and $a=-i$. $(x-i)^2(x+i)^2=(x^2+1)^2$

Mark Bennet
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