Question

If `x=4i` is a root of `f(x) = x^4 - 4x^3 +29x^2 -64x + 208` what are the other roots?

Answer 1

the four roots are:

`+-4i, 2+-3i`

Explanation:

We know that polynomials with real coefficients have roots that must appear in complex conjugate pairs. Therefore, as `4i` is one root then another root must be `-4i`. Subsequently by the factor theorem, then a factor of the `f(x)` is:

`p(x) = (x+4i)(x-4i)`
`\ \ \ \ \ \ \ = (x)^2-(4i)^2`
`\ \ \ \ \ \ \ = x^2+16`

And if we divide `f(x)` by `p(x)` we will be left with a quadratic function, and so we can write:

`f(x) = x^4-4x^3+29x^2-64x+208`
`\ \ \ \ \ \ \ = (x^2+16)(ax^2+bx+c)`

Where `a,b,c` are constant that we must find. If we equate coefficients we get:

`C(x^4) : 1 = a`
`C(x^3) : -4 = b`
`C(x^0) : 208 = 16c => c = 13`

Therefore:

`f(x) = (x^2+16)(x^2-4x+13)`

So the remaining roots are the solution of:

`x^2-4x+13 = 0`
`:. (x-2)^2-2^2+13 = 0`
`:. (x-2)^2+9 = 0`
`:. x-2 = +- 3i`
`:. x = 2 +- 3i`

Hence the four roots are:

`+-4i, 2+-3i`