Aus Online Mathematik Brückenkurs 2

Imagine for a moment taking away all the terms in the numerator apart from \displaystyle x^{3}. If we are to make \displaystyle x^{3} divisible by the denominator \displaystyle x^{2}+3x+1, we need to add and subtract \displaystyle 3x^{2}+x in order to obtain the expression \displaystyle x^{3}+3x^{2}+x=x\left( x^{2}+3x+1 \right),

\displaystyle \begin{align} & \frac{x^{3}+2x^{2}+1}{x^{2}+3x+1}=\frac{x^{3}+\underline{3x^{2}+x-3x^{2}-x}+2x^{2}+1}{x^{2}+3x+1} \\ & =\frac{x^{3}+3x^{2}+x}{x^{2}+3x+1}+\frac{-3x^{2}-x+2x^{2}+1}{x^{2}+3x+1} \\ & =\frac{x\left( x^{2}+3x+1 \right)}{x^{2}+3x+1}+\frac{-x^{2}-x+1}{x^{2}+3x+1} \\ & =x+\frac{-x^{2}-x+1}{x^{2}+3x+1} \\ \end{align}

Now, we carry out the same procedure with the new quotient. To the term \displaystyle -x^{2}, we add and subtract \displaystyle -\text{3}x-\text{1} and obtain

\displaystyle \begin{align} & x+\frac{-x^{2}-x+1}{x^{2}+3x+1}=x+\frac{-x^{2}\underline{-3x-1+3x+1}-x+1}{x^{2}+3x+1} \\ & =x+\frac{-x^{2}-3x-1}{x^{2}+3x+1}+\frac{3x+1-x+1}{x^{2}+3x+1} \\ & =x-1+\frac{2x+2}{x^{2}+3x+1} \\ \end{align}