$\dots +14r+49={(\dots \dots \mathrm{.})}^2\dots \backslash ;+\backslash ;14r+49=\backslash left(\dots \dots .\backslash right)^2$

Am $++$ vor dem Mischterm $14r14r$ erkennt man, dass man die 1. Binomische Formel ${(a+b)}^2=a^2+2ab+b^2\backslash left(a+b\backslash right)^2=a^2+2ab+b^2$ anwenden muss.

Aus dem Quadratterm kann man $bb$ bestimmen.

$49=b^{2}49=b^2$ $\Rightarrow \backslash Rightarrow$ $b=7b=7$

$\dots +14r+49={(\dots +7)}^2\dots \backslash ;+14r+49=\backslash left(\dots \backslash ;+7\backslash right)^2$

Aus dem Mischterm kann man $aa$ bestimmen.

$2\cdot 7\cdot a=14r\Rightarrow a=r2\backslash cdot7\backslash cdot\; a\backslash ;=14r\backslash ;\backslash Rightarrow\backslash ;a=r$

$\Rightarrow {(r+7)}^2\backslash Rightarrow\backslash ;\backslash left(r+7\backslash right)^2$

$r^2-\frac{1}{3}r\dots \mathrm{.}={(\dots \dots )}^{2}r^2-\backslash frac13r\backslash ;\dots .=\backslash left(\dots \dots \backslash right)^2$

Am $--$ vor dem Mischterm $\frac{1}{3}r\backslash frac13r$ erkennt man, dass man die 2. Binomische Formel ${(a+b)}^2=a^2+2ab+b^2\backslash left(a+b\backslash right)^2=a^2+2ab+b^2$ anwenden muss.

Aus Quadratterm kann man $aa$ bestimmen.

$a^2=r^2\Rightarrow a=ra^2=r^2\backslash ;\backslash Rightarrow\backslash ;a=r$

$r^2-\frac{1}{3}r\dots \dots ={(r-\dots )}^{2}r^2-\backslash frac13r\dots \dots \backslash ;=\backslash ;\backslash ;\backslash left(r-\backslash ;\dots \backslash right)^2$

Aus dem Mischterm kann man $bb$ bestimmen.

$2\cdot a\cdot b=\frac{1}{3}r2\backslash cdot\; a\backslash cdot\; b=\backslash frac13r$

$2\cdot r\cdot b=\frac{1}{3}r\Rightarrow b=\frac{1}{6}2\backslash cdot\; r\backslash cdot\; b=\backslash frac13r\backslash ;\backslash Rightarrow\backslash ;b=\backslash frac16$

$\Rightarrow {(r-\frac{1}{6})}^2\backslash Rightarrow\backslash ;\backslash ;\backslash left(r-\backslash frac16\backslash right)^2$