Find $p(x)$ from the sum of $p(x+a)$ and $ p(x+b)$ asked by Katrine Linda in the pretty math group

Exercise:

Determine $p\left( x \right)$ such that $p\left( x-2 \right)+p\left( x+1 \right)=2{{x}^{2}}-4x+2$

Solution : we have $p\left( x-2 \right)\,\And p\left( x+1 \right)\in \mathbb{R}\left[ x \right]$ of degree two both thus its sum must be of

degree 2 hence $p\left( x-2 \right)=a{{\left( x-2 \right)}^{2}}+b\left( x-2 \right)+c$ and $p\left( x+1 \right)=a{{\left( x+1 \right)}^{2}}+b\left( x+1 \right)+d$

Where $a,b,c,d\in \mathbb{R}$

So $p\left( x-2 \right)+p\left( x+1 \right)=a\left( {{x}^{2}}-4x+4+{{x}^{2}}+2x+1 \right)+b\left( x-2+x+1 \right)+c+d=2{{x}^{2}}-4x+2$

$\Leftrightarrow a\left( 2{{x}^{2}}-2x+5 \right)+b\left( 2x-1 \right)+c+d=2{{x}^{2}}-4x+2$

$\Leftrightarrow 2a{{x}^{2}}+x\left( 2b-2a \right)+5-b+c+d=2{{x}^{2}}-4x+2$

$\Leftrightarrow \left\{ \begin{align}
  & 2a=2 \\
 & 2b-2a=-4 \\
 & 5-b+c+d=2 \\
\end{align} \right.$

So $a=1\,\,,b=-1$ and $c+d=2-6=-4$ thus $c=d=-2$ Hence $p\left( x \right)={{x}^{2}}-x-2$