Question

Identify all the roots.? `x^4 - x^3 + 3x^2 - 9x - 54 = 0`

Answer 1

The roots are `-2`, `3` and `+-3i`

Explanation:

Given:

`x^4-x^3+3x^2-9x-54=0`

The rational roots theorem tells us that any rational roots are expressible in the form `p/q` for integers `p, q` with `p` a divisor of the constant term `-54` and `q` a divisor of the coefficient `1` of the leading term.

That means that the only possible rational roots are:

`+-1, +-2, +-3, +-6, +-9, +-18, +-27, +-54`

Note also that the pattern of signs of the coefficients is `+ - + - -`. With `3` changes of sign, Descartes' Rule of Signs tells us that there could be `1` or `3` positive real roots.

Reversing the sign of the coefficients on the terms of odd degree we get the pattern `+ + + + -`. With `1` change of sign, Descartes' Rule of Signs tells us that this quartic equation has exactly `1` negative real root.

Trying some values, we find:

`(color(blue)(-2))^4-(color(blue)(-2))^3+3(color(blue)(-2))^2-9(color(blue)(-2))-54 = 16+8+12+18-54 = 0`

So `x=-2` is a root and `(x+2)` a factor:

`x^4-x^3+3x^2-9x-54=(x+2)(x^3-3x^2+9x-27)`

Note that in the remaining cubic factor, the ratio of the first and second terms is the same as that of the third and fourth terms. So this cubic will factor by grouping:

`x^3-3x^2+9x-27 = (x^3-3x^2)+(9x-27)`

`color(white)(x^3-3x^2+9x-27) = x^2(x-3)+9(x-3)`

`color(white)(x^3-3x^2+9x-27) = (x^2+9)(x-3)`

The remaining quadratic `x^2+9` has only complex zeros:

`x^2+9 = x^2-9i^2 = x^2-(3i)^2 = (x-3i)(x+3i)`

So the roots are `-2`, `3` and `+-3i`