How do you find all the real and complex roots of x^4 + 7x^3 + 31x^2 + 175x + 150 = 0?

Feb 23, 2016

Use the rational root theorem to find the two Real roots $- 1$, $- 6$ and factor the quartic to find the remaining quadratic factor and Complex roots $\pm 5 i$

Explanation:

$f \left(x\right) = {x}^{4} + 7 {x}^{3} + 31 {x}^{2} + 175 x + 150$

By the rational root theorem, any rational roots of $f \left(x\right) = 0$ must be expressible in the form $\frac{p}{q}$ where $p$ and $q$ are integers, $p$ a divisor of the constant term $150$ and $q$ a divisor of the coefficient $1$ of the leading term.

In addition, since all of the coefficients of $f \left(x\right)$ are positive, it has no positive roots.

So the only possible rational roots are:

$- 1 , - 2 , - 3 , - 5 , - 6 , - 10 , - 15 , - 25 , - 30 , - 50 , - 75 , - 150$

Trying each in turn, we find:

$f \left(- 1\right) = 1 - 7 + 31 - 175 + 150 = 0$
...
$f \left(- 6\right) = 1296 - 1512 + 1116 - 1050 + 150 = 0$

So $x = - 1$ and $x = - 6$ are roots and $\left(x + 1\right)$ and $\left(x + 6\right)$ are factors of $f \left(x\right)$

${x}^{4} + 7 {x}^{3} + 31 {x}^{2} + 175 x + 150$

$= \left(x + 1\right) \left({x}^{3} + 6 {x}^{2} + 25 x + 150\right)$

$= \left(x + 1\right) \left(x + 6\right) \left({x}^{2} + 25\right)$

The remaining quadratic factor has zeros $\pm 5 i$

So the roots of $f \left(x\right) = 0$ are $- 1$, $- 6$, $\pm 5 i$