de-CH
utf-8
math math-format
Polynomial Division
g-06-01
multiple
65536
randRangeNonZero( -8, 8 ) randRangeNonZero( -8, 8 )
randRangeNonZero(-8,8) randRangeNonZero(-8,8) randRangeNonZero(-8,8)

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.