Question

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

Answer 1

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 `+-5i`

Explanation:

`f(x) = x^4+7x^3+31x^2+175x+150`

By the rational root theorem, any rational roots of `f(x) = 0` must be expressible in the form `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(x)` 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(-1) = 1-7+31-175+150 = 0`
...
`f(-6) = 1296-1512+1116-1050+150 = 0`

So `x=-1` and `x=-6` are roots and `(x+1)` and `(x+6)` are factors of `f(x)`

`x^4+7x^3+31x^2+175x+150`

`=(x+1)(x^3+6x^2+25x+150)`

`=(x+1)(x+6)(x^2+25)`

The remaining quadratic factor has zeros `+-5i`

So the roots of `f(x) = 0` are `-1`, `-6`, `+-5i`