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
$\left(x^3-9\right)\left(2\cdot x^2-4\right)=0$
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.

$x=9^{{{1}\over{3}}} \lor x=-\sqrt{2} \lor x=\sqrt{2}$

$\begin{array}{rcl} \left(x^3-9\right)\left(2\cdot x^2-4\right)&=&0 \\ &&\phantom{xxx}\blue{\text{the equation we need to solve}} \\ x^3-9=0 &\lor& 2\cdot x^2-4=0 \\ &&\phantom{xxx}\blue{A \cdot B =0 \Leftrightarrow A=0 \lor B=0} \\ x^3=9 &\lor& 2\cdot x^2=4 \\ &&\phantom{xxx}\blue{\text{constants moved to the right hand side}} \\ \displaystyle x^3=9 &\lor& \displaystyle x^2=\displaystyle 2 \\ &&\phantom{xxx}\blue{\text{divided by coefficient in front of term with }x} \\ \displaystyle x=9^{{{1}\over{3}}} &\lor& \displaystyle x=-\sqrt{2} \lor x=\sqrt{2} \\ &&\phantom{xxx}\blue{\text{extracted the root}} \\ \end{array}$

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