# How do you factor completely 2x^3+10x^2+14x+70?

Jan 4, 2017

$2 {x}^{3} + 10 {x}^{2} + 14 x + 70 = 2 \left({x}^{2} + 7\right) \left(x + 5\right)$

#### Explanation:

Note that the ratio of the first and second terms is the same as that between the third and fourth terms. So this cubic will factor by grouping, but first we can separate out the common scalar factor $2$...

$2 {x}^{3} + 10 {x}^{2} + 14 x + 70 = 2 \left({x}^{3} + 5 {x}^{2} + 7 x + 35\right)$

$\textcolor{w h i t e}{2 {x}^{3} + 10 {x}^{2} + 14 x + 70} = 2 \left(\left({x}^{3} + 5 {x}^{2}\right) + \left(7 x + 35\right)\right)$

$\textcolor{w h i t e}{2 {x}^{3} + 10 {x}^{2} + 14 x + 70} = 2 \left({x}^{2} \left(x + 5\right) + 7 \left(x + 5\right)\right)$

$\textcolor{w h i t e}{2 {x}^{3} + 10 {x}^{2} + 14 x + 70} = 2 \left({x}^{2} + 7\right) \left(x + 5\right)$

That's as far as we can go with Real coefficients.

If you allow Complex coefficients then:

${x}^{2} + 7 = \left(x - \sqrt{7} i\right) \left(x + \sqrt{7} i\right)$