Polynomial equation Is it possible to find polynomials with rational coefficients $P(x),Q(x)$ such that $Px^3+Px^2+Qx+2Q=1$? I have trying in vain to find one by inspection, but that might just be me.
 A: Write the equation as 
$$ P(x^3+x^2) + Q(x+2) = 1. $$
Since $\text{gcd}(x^3+x^2,x+2)=1$ over $\mathbb{Q}$ the solution exists and can be found by employing the Euclidean algorithm. 
A: Just in case the OP does not know the Euclidean algorithm, I'll compute it.
Divide $(x^3 + x^2)$ by $x+2$. This is just ordinary high school long division. So we find that
$$\begin{equation}(x^3 + x^2) = (x+2)q(x) + r(x)\end{equation}$$
where $q(x) = x^2 - x$ and $r(x) = 2x$. Now by the Euclidean algorithm we then divide $(x+ 2)$ by $2x$ to get
$(x+2) = 2x(\frac{1}{2}) + 2$ so that $2 = (x+2) - 2x(\frac{1}{2})$. Now from the first expression we found that $2x = (x^3 + x^2) - (x+2)(x^2 - x)$, so that substituting it in here we get
$$\begin{eqnarray*}2 &=& (x+2) - \frac{\bigg[(x^3 + x^2) - (x+2)(x^2 - x)         \bigg]}{2}\\
&=& (x+2)\frac{(2+ (x^2 - x))}{4} - \frac{[x^3 + x^2]}{4} \end{eqnarray*} $$
Reading off this your $P$ is $-1/4$ while $$ Q= \frac{(2+ (x^2 - x))}{4} .$$
A: Yes, by the division  algorithm in $\rm\:\mathbb Q[x],\:$ since $\rm\:x^3+x^2,\ x+2\:$ are coprime, the extended Euclidean algorithm yields a Bezout identity for their gcd: $\rm\: g(x)\:(x^3+x^2) + h(x)\:(x+2) = 1,\:$ some $\rm\:g,h\in \mathbb Q[x].\:$ When, as here, one polynomial has degree one, the Euclidean algorithm has only one step, yielding the well-known Remainder Theorem
$$\rm f(a)\ =\:\ f(x)\:\ -\ \:\frac{f(x)-f(a)}{x-a}\:(x-a)$$
$$\begin{eqnarray}\rm -4\ &=\rm\ x^3+x^2\: -\: \frac{x^3+x^2+4}{x+2}(x+2) \\  &=\rm\ x^3+x^2\: -\: (x^2-x+2)\:(x+2)\end{eqnarray}$$
Multiplying the above by $-1/4\:$ yields the desired Bezout identity.
