6.3 The greatest common factor

Factor $3x-18.$

The greatest common factor is 3.

$\begin{array}{lll}3x-18=3\cdot x-3\cdot 6\hfill & \hfill & \text{Factor}\text{out}3.\hfill \\ 3x-18=3()\hfill & \hfill & \text{Divide}\text{each}\text{term}\text{of}\text{the}\text{product}\text{by}3.\hfill \\ \hfill & \hfill & \frac{3x}{3}=x\text{and}\frac{-18}{3}=-6\hfill \\ \hfill & \hfill & (\text{Try}\text{to}\text{perform}\text{this}\text{division}\text{mentally}\text{.})\hfill \\ 3x-18=3(x-6)\hfill & \hfill & \hfill \end{array}$

Factor $9x^3+18x^2+27x.$

Notice that $9x$ is the greatest common factor.

$\begin{array}{ll}9x^3+18x^2+27x=9x\cdot x^2+9x\cdot 2x+9x\cdot 3.\hfill & \text{Factor}\text{out}9x.\hfill \\ 9x^{\text{3}}+18x^2+27x=9x()\hfill & \text{Mentally}\text{divide}9x\text{into}\text{each}\text{term}\text{of}\text{the}\text{product}\text{.}\hfill \\ \text{9}{x}^3+18x^2+27x=9x(x^2+2x+3)\hfill & \hfill \end{array}$

Factor $10x^2{y}^3-20x{y}^4-35y^5.$

Notice that $5y^3$ is the greatest common factor. Factor out $5y^3.$

$10x^2{y}^3-20x{y}^4-35y^5=5y^3()$

Mentally divide $5y^3$ into each term of the product and place the resulting quotients inside the $().$

$10x^2{y}^3-20x{y}^4-35y^5=5y^3(2x^2-4xy-7y^2)$

Factor $-12x^5+8x^3-4x^2.$

We see that the greatest common factor is $-4x^2.$

$-12x^5+8x^3-4x^2=-4x^2()$

Mentally dividing $-4x^2$ into each term of the product, we get

$-12x^5+8x^3-4x^2=-4x^2(3x^3-2x+1)$

Factor $(x-7)a+(x-7)b.$

Notice that $(x-7)$ is the greatest common factor. Factor out $(x-7).$

$\begin{array}{ccc}(x-7)a+(x-7)b& =& (x-7)(\begin{array}{cc}& \end{array})\\ \text{Then},\frac{(x-7)a}{(x-7)}& =& a\text{and}\frac{(x-7)b}{(x-7)}=b.\\ (x-7)a+(x-7)b& =& (x-7)(a+b)\end{array}$

Factor $3x^2(x+1)-5x(x+1)$ .

Notice that $x$ and $(x+1)$ are common to both terms. Factor them out. We’ll perform this factorization by letting $A=x(x+1).$ Then we have

$\begin{array}{ccc}3xA-5A& =& A(3x-5)\\ \text{But}A& =& x(x+1),\text{so}\\ 3x^2(x+1)-5x(x+1)& =& x(x+1)(3x-5)\end{array}$