# Factor completely (linear factors only): f(x)=3x^4+10x^3+24x^2+22x+5?

Oct 5, 2017

=$\left(x + 1\right) \left(3 x + 1\right) \left(x + 1 + 2 i\right) \left(x + 1 - 2 i\right)$

#### Explanation:

$3 {x}^{4} + 10 {x}^{3} + 24 {x}^{2} + 22 x + 5$

=$3 {x}^{4} + 3 {x}^{3} + 7 {x}^{3} + 7 {x}^{2} + 17 {x}^{2} + 17 x + 5 x + 5$

=$3 {x}^{3} \cdot \left(x + 1\right) + 7 {x}^{2} \cdot \left(x + 1\right) + 17 x \cdot \left(x + 1\right) + 5 \cdot \left(x + 1\right)$

=$\left(x + 1\right) \cdot \left(3 {x}^{3} + 7 {x}^{2} + 17 x + 5\right)$

=$\left(x + 1\right) \cdot \left(3 {x}^{3} + {x}^{2} + 6 {x}^{2} + 2 x + 15 x + 5\right)$

=$\left(x + 1\right) \cdot \left[{x}^{2} \left(3 x + 1\right) + 2 x \left(3 x + 1\right) + 5 \left(3 x + 1\right)\right]$

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

=$\left(x + 1\right) \left(3 x + 1\right) \left[{x}^{2} + 2 x + 1 - \left(- 4\right)\right]$

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

=$\left(x + 1\right) \left(3 x + 1\right) \left(x + 1 + 2 i\right) \left(x + 1 - 2 i\right)$