# How do you factor 40x^3 + 5?

Jan 1, 2016

$5 \left(8 {x}^{3} + 1\right)$

#### Explanation:

In the expression $40 {x}^{3} + 5$ the only common factor is 5

So we have factored this expression as

$5 \left(8 {x}^{3} + 1\right)$

Nov 2, 2017

$5 \left(2 x + 1\right) \left(4 {x}^{2} - 2 x + 1\right)$

#### Explanation:

$\text{take out a "color(blue)"common factor of 5}$

$\Rightarrow 40 {x}^{3} + 5$

$= 5 \left(8 {x}^{3} + 1\right)$

$8 {x}^{3} + 1 \text{ is a "color(blue)"sum of cubes}$

•color(white)(x)a^3+b^3=(a+b)(a^2-ab+b^2)

$\text{note that "8x^3=(2x)^3" and } 1 = {1}^{3}$

$\text{here "a=2x" and } b = 1$

$\Rightarrow 8 {x}^{3} + 1 = \left(2 x + 1\right) \left({\left(2 x\right)}^{2} - \left(2 x .1\right) + {1}^{2}\right)$

$\textcolor{w h i t e}{\Rightarrow 8 {x}^{3} + 1} = \left(2 x + 1\right) \left(4 {x}^{2} - 2 x + 1\right)$

$\Rightarrow 40 {x}^{3} + 5 = 5 \left(2 x + 1\right) \left(4 {x}^{2} - 2 x + 1\right)$