How do you factor the expression (2x-3)^3 (x+1) +(x-3) (2x-3)^2?

Jan 12, 2016

Combine, simplify and use the difference of squares identity to find:

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

$= 2 {\left(2 x - 3\right)}^{2} \left(x - \sqrt{3}\right) \left(x + \sqrt{3}\right)$

Explanation:

The difference of squares identity can be written:

${a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)$

We use this with $a = x$ and $b = \sqrt{3}$ below...

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

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

$= {\left(2 x - 3\right)}^{2} \left(\left(2 {x}^{2} - x - 3\right) + \left(x - 3\right)\right)$

$= {\left(2 x - 3\right)}^{2} \left(2 {x}^{2} - 6\right)$

$= 2 {\left(2 x - 3\right)}^{2} \left({x}^{2} - 3\right)$

$= 2 {\left(2 x - 3\right)}^{2} \left({x}^{2} - {\left(\sqrt{3}\right)}^{2}\right)$

$= 2 {\left(2 x - 3\right)}^{2} \left(x - \sqrt{3}\right) \left(x + \sqrt{3}\right)$