Functions: Higher degree polynomials
Equations with polynomials
Equations with higher degree polynomials are not always solvable by hand. But here we will take a look at some situations in which it is easily possible.
\[\blue A \cdot \green B=0\]
gives
\[\blue A=0 \lor \green B=0\]
Example
\[\begin{array}{c}\blue{x^2}\left(\green{x^2-2}\right)=0 \\ \\ \blue{x^2}=0 \lor \green{x^2-2}=0 \end{array}\]
\[\blue{A}^2=\green{B}^2\]
gives
\[\blue{A}=\green{B} \lor \blue{A}=-\green{B}\]
Example
\[\begin{array}{c}\left(\blue{x^2-4x}\right)^2=\green{x}^2 \\ \\ \blue{x^2-4x}=\green{x} \lor \blue{x^2-4x}=-\green{x} \end{array}\]
\[\blue{A} \cdot \green{B} = \blue{A} \cdot \purple{C}\]
gives
\[\blue{A}=0 \lor \green{B}=\purple{C}\]
Example
\[\begin{array}{c}\left(\blue{x^2-4}\right)\cdot\green{x^2}=\left(\blue{x^2-4}\right)\cdot\left(\purple{6x-8}\right) \\\\ \blue{x^2-4}=0 \lor \green{x^2} = \purple{6x-8}\end{array}\]
\[\blue{A} \cdot \green{B} = \blue{A} \]
gives
\[\blue{A}=0 \lor \green{B}=1\]
Example
\[\begin{array}{c}\left(\blue{x^2-4}\right)\cdot\green{x^2}=\left(\blue{x^2-4}\right)\ \\ \\ \blue{x^2-4}=0 \lor \green{x^2} = 1\end{array}\]
#\begin{array}{rcl}
\left(x^3-9\right)\left(6-4\cdot x^2\right)&=&0 \\ &&\phantom{xxx}\blue{\text{the equation we need to solve}} \\
x^3-9=0 &\lor& 6-4\cdot x^2=0 \\ &&\phantom{xxx}\blue{A \cdot B =0 \Leftrightarrow A=0 \lor B=0} \\
x^3=9 &\lor& -4\cdot x^2=-6 \\ &&\phantom{xxx}\blue{\text{constants moved to the right hand side}} \\
x^3=9 &\lor& x^2={{3}\over{2}} \\ &&\phantom{xxx}\blue{\text{divided by coefficient in front of term with }x} \\
x=9^{{{1}\over{3}}} &\lor& x=-{{\sqrt{3}}\over{\sqrt{2}}} \lor x={{\sqrt{3}}\over{\sqrt{2}}}
\\ &&\phantom{xxx}\blue{\text{extracted the root}} \\
\end{array}#
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