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 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 $4x^2 \cdot (x+2)$.