We can factor out common factors:
\[\begin{array}{rcl}\blue{a}\green{b}+\blue{a}\purple{c} &=& \blue{a}(\green{b}+\purple{c}) \\ \\ \blue{a}\purple{c}+\green{b}\purple{c}&=&(\blue{a}+\green{b})\purple{c}\end{array}\]
The examples show that the common factor can be a number or an expression with a variable.
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Examples
\[\begin{array}{rcl}3x+27&=&\blue{3} \cdot \green{x}+\blue{3}\cdot \purple{9}\\&=&\blue{3} \cdot (\green{x}+\purple{9})\\ \\ 3x^2+x &=& \blue{x} \cdot \green{3x} + \blue{x} \cdot \purple{1} \\&=& \blue{x} \cdot (\green{3x}+ \purple{1})\\ \\ 4x^2+8x&=& \blue{4x} \cdot \green{x} + \blue{4x} \cdot \purple{2} \\&=& \blue{4x} \cdot (\green{x}+ \purple{2})\end{array}\]
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The above rule also applies to multiple terms within the brackets:
\[\begin{array}{rcl}\blue{a} \cdot \green{b} + \blue{a} \cdot \purple{c} + \blue{a} \cdot \color{purple}{d}&=&\blue{a} \cdot \left(\green{b}+\purple{c}+\color{purple}{d}\right) \\ \\
\blue{a} \cdot \green{b} + \blue{a} \cdot \purple{c} + \blue{a} \cdot \color{purple}{d}+\blue{a} \cdot \orange{e}&=&\blue{a} \cdot \left(\green{b}+\purple{c}+\color{purple}{d}+\orange{e}\right)\end{array}\]
Examples
\[\begin{array}{rcl}2x^3+4x^2+6x&=&\blue{2x} \cdot \green{x^2} + \blue{2x} \cdot \purple{2x} + \blue{2x} \cdot \color{purple}{3}\\&=&\blue{2x} \cdot \left(\green{x^2}+\purple{2x}+\color{purple}{3}\right) \\ \\ 4x^4+8x^3+4x^2+12x&=&\blue{4x} \cdot \green{x^3} + \blue{4x} \cdot \purple{2x^2} + \blue{4x} \cdot \color{purple}{x}+\blue{4x} \cdot \orange{3}\\&=&\blue{4x} \cdot \left(\green{x^3}+\purple{2x^2}+\color{purple}{x}+\orange{3}\right)\end{array}\]
In the above rule, there are often multiple options for factoring out factors. If we factor out a factor #4#, we can also factor out a factor #2#. In general, we're interested in factoring out the largest possible factor.
If we have factored out as many factors as possible, we call it factorization. The process of factoring out is called factorization.
In the examples below, we will see that we can sometimes factor out multiple factors. We will call the last step in factoring out 'factorization', because we will have factored out the largest common term.
\[\begin{array}{rcl} 4x^2+2x &=& 2 \cdot (x^2 +x) \\ &=& 2x \cdot (x +1) \end{array}\]
The largest factor we have factored out in the example above is #2x#. The factorization is therefore #2x \cdot (x +1)#.
\[\begin{array}{rcl} 4x^3+8x^2 &=& 4 \cdot (x^3 +2x^2) \\ &=& 4x \cdot (x^2 +2x) \\ &=& 4x^2 \cdot (x+2) \end{array}\]
The largest factor we have factored out in the example above is #4x^2#. The factorization is therefore #4x^2 \cdot (x+2)#.
Except for the order of the terms and the multiplication of constants, the factorization is unique if there are no other factors to factor out.
At the moment, we only work with integers and fractions and not with roots.
If possible, factorize: #2\cdot a-18#.
#2\cdot \left(a-9\right)#
When factorizing, we look at the greatest common factor to both terms. In this case, the greatest common factor is #2#.
\[\begin{array}{rcl}
2\cdot a-18 &=& 2 \cdot a + 2 \cdot -9
\\ &&\phantom{xxx}\blue{\text{written with the common factor }}\\
&=& 2\cdot \left(a-9\right)
\\ &&\phantom{xxx}\blue{\text{factorized }}\\
\end{array}\]