# How do you factor 21x^3 - 18x^2 + 7x - 6 by grouping?

Jun 29, 2016

Expression$= 3 {x}^{2} \left(7 x - 6\right) + 1 \left(7 x - 6\right) = \left(7 x - 6\right) \left(3 {x}^{2} + 1\right) .$

#### Explanation:

Observe that $3 {x}^{2}$ can be taken out as common from the first two terms, 7, from the last two terms, only $1$ can be taken out. Hence,

Expression$= 3 {x}^{2} \left(7 x - 6\right) + 1 \left(7 x - 6\right) = \left(7 x - 6\right) \left(3 {x}^{2} + 1\right) .$

Jun 29, 2016

$\left(7 x - 6\right) \left(3 {x}^{2} + 1\right)$

#### Explanation:

Group the terms in 'pairs' as follows.

$\left[21 {x}^{3} - 18 {x}^{2}\right] + \left[7 x - 6\right]$

now factorise each 'pair'.

$\textcolor{red}{3 {x}^{2}} \left(7 x - 6\right) \textcolor{red}{+ 1} \left(7 x - 6\right)$

We now have a common factor of (7x -6) in each pair which we can take out.

$\left(7 x - 6\right) \left(\textcolor{red}{3 {x}^{2} + 1}\right)$

$\Rightarrow 21 {x}^{3} - 18 {x}^{2} + 7 x - 6 = \left(7 x - 6\right) \left(3 {x}^{2} + 1\right)$