# How do you find all the real and complex roots of 2x^4 - 5x^3 + 53x^2 - 125x + 75 = 0?

Jan 14, 2016

Check the sum of the coefficients to find root $x = 1$, then divide by $\left(x - 1\right)$ and factor by grouping to find the remaining roots $x = \frac{3}{2}$ and $x = \pm 5 i$.

#### Explanation:

$f \left(x\right) = 2 {x}^{4} - 5 {x}^{3} + 53 {x}^{2} - 125 x + 75$

First note that the sum of the coefficients is zero.

That is: $2 - 5 + 53 - 125 + 75 = 0$

So $f \left(1\right) = 0$ and $\left(x - 1\right)$ is a factor.

Use your favourite method to divide $f \left(x\right)$ by $\left(x - 1\right)$. Here I long divide the coefficients: Then factor by grouping...

$f \left(x\right) = \left(x - 1\right) \left(2 {x}^{3} - 3 {x}^{2} + 50 x - 75\right)$

$= \left(x - 1\right) \left(\left(2 {x}^{3} - 3 {x}^{2}\right) + \left(50 x - 75\right)\right)$

$= \left(x - 1\right) \left({x}^{2} \left(2 x - 3\right) + 25 \left(2 x - 3\right)\right)$

$= \left(x - 1\right) \left({x}^{2} + 25\right) \left(2 x - 3\right)$

The remaining quadratic factor has no linear factors with Real coefficients since ${x}^{2} + 25 \ge 25 \ne 0$ for any Real values of $x$, but it does have zeros $\pm 5 i$, since ${\left(5 i\right)}^{2} = {\left(- 5 i\right)}^{2} = - 25$

$= \left(x - 1\right) \left(x - 5 i\right) \left(x + 5 i\right) \left(2 x - 3\right)$

So $f \left(x\right)$ has zeros $x = 1$, $x = \frac{3}{2}$, $x = \pm 5 i$