Determine {\color{blue}a}, {\color{red}b} and {\color{orange}c} such that:
\left(
S*A x^4 + L*A x^3 + C*A+S*B x^2 + L*B x + C*B
\right) : {\color{teal}\left( A x^2 + 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 choose {\color{blue}a} such that
{\color{blue}a}x^2 \cdot {\color{teal}negParens(A)x^2}=
S*Ax^4 holds:
Thus {\color{blue}a} =S.
For this {\color{blue}a} we multiply {\color{blue}S x^2} and
{\color{teal}\left( A x^2 + B \right ) } and get
S*A x^4 + S*B {x^2} .
It is S*A x^4 + L*A x^3 +C*A+S*B x^2 -
\left ( S*A x^4 + S*B x^2
\right )
= L*A x^3 + C*A x^2 , and we need to choose
{\color{red}b} such that {\color{red}bx} \cdot {\color{teal}negParens(A)x^2} =
{\color{red}L*A x^3}.
Thus {\color{red}b} =L.
With this {\color{red}b} we multiply {\color{red}bx} and
{\color{teal}\left( A x^2 + B \right ) } and get
L*A x^3 + L*B x .
We subtract again
L*A x^3 + C*A x^2 + L*Bx - \left ( L*A x^3 + L*Bx
\right )
= {\color{orange}C*A x^2}, and we need to choose
{\color{orange}c} such that {\color{orange}c} \cdot {\color{teal}negParens(A)x^2} =
{\color{orange}C*A x^2}.
Thus {\color{orange}c} =C.
The calculation works, as we see
{\color{orange}c} \cdot {\color{teal}\left( A x^2 + B \right ) } =
C*A x^2 + C*B und {\color{orange}C*A x^2} + C*B -
\left( C*A x^2 + C*B \right)
=0.