# How do you find all the real and complex roots of f(x)= 3x^3 + 36x^2 + 156x + 240?

Dec 11, 2016

$f \left(x\right)$ has zeros $- 4$ and $- 4 \pm 2 i$

#### Explanation:

The difference of squares identity can be written:

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

We use this later, with $a = \left(x + 4\right)$ and $b = 2 i$

$f \left(x\right) = 3 {x}^{3} + 36 {x}^{2} + 156 x + 240$

$\textcolor{w h i t e}{f \left(x\right)} = 3 \left({x}^{3} + 12 {x}^{2} + 52 x + 80\right)$

We can start to simplify this cubic using a linear substitution to eliminate the square term:

${x}^{3} + 12 {x}^{2} + 52 x + 80 = \left({x}^{3} + 12 {x}^{2} + 48 x + 64\right) + \left(4 x + 16\right)$

$\textcolor{w h i t e}{{x}^{3} + 12 {x}^{2} + 52 x + 80} = {\left(x + 4\right)}^{3} + 4 \left(x + 4\right)$

We could put $t = x + 4$ at this stage, but there is no constant term, so let's leave $\left(x + 4\right)$ as $\left(x + 4\right)$ and just factor it from here:

$\textcolor{w h i t e}{{x}^{3} + 12 {x}^{2} + 52 x + 80} = \left(x + 4\right) \left({\left(x + 4\right)}^{2} + 4\right)$

$\textcolor{w h i t e}{{x}^{3} + 12 {x}^{2} + 52 x + 80} = \left(x + 4\right) \left({\left(x + 4\right)}^{2} - {\left(2 i\right)}^{2}\right)$

Having made the quadratic factor into a difference of squares, we can apply the difference of squares identity to find:

$\textcolor{w h i t e}{{x}^{3} + 12 {x}^{2} + 52 x + 80} = \left(x + 4\right) \left(\left(x + 4\right) - 2 i\right) \left(\left(x + 4\right) + 2 i\right)$

$\textcolor{w h i t e}{{x}^{3} + 12 {x}^{2} + 52 x + 80} = \left(x + 4\right) \left(x + 4 - 2 i\right) \left(x + 4 + 2 i\right)$

Hence the zeros of $f \left(x\right)$ are $- 4$ and $- 4 \pm 2 i$