Equations with polynomials

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

\[\begin{array}{c}\left(\blue{x^2-4x}\right)^2=\green{x}^2 \\ \\ \blue{x^2-4x}=\green{x} \lor \blue{x^2-4x}=-\green{x} \end{array}\]

We can prove this rule by means of reduction, factorization and the rule #A \cdot B=0# if and only if #A=0 \lor B=0#. This is done in the following manner:

\[\begin{array}{rcl}A^2&=&B^2 \\ &&\phantom{xxx}\blue{\text{the equation we need to solve}} \\ A^2-B^2&=&0 \\ &&\phantom{xxx}\blue{\text{both sides minus }B^2} \\ \left(A-B\right)\left(A+B\right)&=&0 \\ &&\phantom{xxx}\blue{\text{factorized}} \\ A-B=0 &\lor& A+B=0 \\ &&\phantom{xxx}\blue{A\cdot B=0 \Leftrightarrow A=0 \lor B=0} \\ A=B &\lor& A=-B \\ &&\phantom{xxx}\blue{B \text{ moved to the other side}} \end{array}\]

\[\blue{A} \cdot \green{B} = \blue{A} \cdot \purple{C}\]

gives

\[\blue{A}=0 \lor \green{B}=\purple{C}\]

Example

\[\begin{array}{c}\left(\blue{x^2-4}\right)\cdot\green{x^2}=\left(\blue{x^2-4}\right)\cdot\left(\purple{6x-8}\right) \\\\ \blue{x^2-4}=0 \lor \green{x^2} = \purple{6x-8}\end{array}\]

We can prove this rule by means of reduction, factorization and the rule #A \cdot B=0# if and only if #A=0 \lor B=0#. This is done in the following manner:

\[\begin{array}{rcl}A \cdot B&=&A \cdot C \\ &&\phantom{xxx}\blue{\text{the equation we need to solve}} \\ A \cdot B - A \cdot C&=&0 \\ &&\phantom{xxx}\blue{\text{both sides minus }A \cdot C} \\ A\cdot \left(B-C\right)&=&0 \\ &&\phantom{xxx}\blue{\text{factorized}} \\ A=0 &\lor& B-C=0 \\ &&\phantom{xxx}\blue{A\cdot B=0 \Leftrightarrow A=0 \lor B=0} \\ A=0 &\lor& B=C\\ &&\phantom{xxx}\blue{C \text{ moved to the other side}} \end{array}\]

\[\blue{A} \cdot \green{B} = \blue{A} \]

gives

\[\blue{A}=0 \lor \green{B}=1\]

Example

\[\begin{array}{c}\left(\blue{x^2-4}\right)\cdot\green{x^2}=\left(\blue{x^2-4}\right)\ \\ \\ \blue{x^2-4}=0 \lor \green{x^2} = 1\end{array}\]

This can be proven easily by choosing #\purple{C}=1# in the previous rule

\[\blue{A} \cdot \green{B} = \blue{A} \cdot \purple{C} \Leftrightarrow \blue{A}=0 \lor \green{B}=\purple{C}\]

Solve for #x# in
\[\left(x^3-6\right)\left(5\cdot x^2-2\right)=0\]
Give your answer in the form #x=x_1 \lor x=x_2 \lor \ldots \lor x=x_n#, where #x_1#, #x_2#, #\ldots#, #x_n# are exact numbers (no decimal values), and #n# the number of solutions.

# x=6^{{{1}\over{3}}} \lor x=-{{\sqrt{2}}\over{\sqrt{5}}} \lor x={{\sqrt{2}}\over{\sqrt{5}}} #

#\begin{array}{rcl}
\left(x^3-6\right)\left(5\cdot x^2-2\right)&=&0 \\ &&\phantom{xxx}\blue{\text{the equation we need to solve}} \\
x^3-6=0 &\lor& 5\cdot x^2-2=0 \\ &&\phantom{xxx}\blue{A \cdot B =0 \Leftrightarrow A=0 \lor B=0} \\
x^3=6 &\lor& 5\cdot x^2=2 \\ &&\phantom{xxx}\blue{\text{constants moved to the right hand side}} \\
x^3=6 &\lor& x^2={{2}\over{5}} \\ &&\phantom{xxx}\blue{\text{divided by coefficient in front of term with }x} \\
x=6^{{{1}\over{3}}} &\lor& x=-{{\sqrt{2}}\over{\sqrt{5}}} \lor x={{\sqrt{2}}\over{\sqrt{5}}}
\\ &&\phantom{xxx}\blue{\text{extracted the root}} \\
\end{array}#

New example