Determine {\color{blue}a}, {\color{red}b}
and {\color{orange}c}
such that:
\left(
S*A x^3 + L*A+S*B x^2 + C*A+L*B x + C*B
\right) : {\color{teal}\left( A x + B \right ) }=
{\color{blue}a} x^2 + {\color{red}b} x + {\color{orange}c}
.
{\color{blue}a}
=
S
{\color{red}b}
=
L
{\color{orange}c}
=
C
We need to find {\color{blue}a}
such that
{\color{blue}a}x^2 \cdot {\color{teal}negParens(A)x} = S*Ax^3
holds:
Thus {\color{blue}a} =S
.
For this {\color{blue}a}
we multiply
{\color{blue}S x^2}
and
{\color{teal}\left( A x + B \right ) }
and get
S*A x^3 + S*B {\color{red}x^2}
.
It is S*A x^3 + L*A+S*B x^2 - \left ( S*A x^3 + S*B {\color{red}x^2}
\right )
= {\color{red}L*A x^2}
, and we need to choose
{\color{red}b}
such that {\color{red}bx} \cdot {\color{teal}negParens(A)x}=
{\color{red}L*A x^2}
.
Thus {\color{red}b} =L
.
With this {\color{red}b}
we multiply {\color{red}bx}
and
{\color{teal}\left( A x + B \right ) }
and get
L*A x^2 + L*B {\color{orange}x}
.
We subtract again
L*A x^2 + C*A+L*B x - \left ( L*A x^2 + L*B {\color{orange}x}
\right )
= {\color{orange}C*A x}
, and we need to choose
{\color{orange}c}
such that {\color{orange}c} \cdot {\color{teal}negParens(A)x}=
{\color{orange}C*A x}
.
This {\color{orange}c} =C
.
The calculation works, as we see
{\color{orange}c} \cdot {\color{teal}\left( A x + B \right ) } =
C*A x + C*B
und {\color{orange}C*A x}+ C*B -
\left( C*A x + C*B \right)
=0.