Question

If 2-3i is a root of `z^3-7z^2+25z-39=0`, find the other two roots?

Answer 1

2+3i, and 3.

Explanation:

I polynomial equations of order 3 (cubics like this) written in the form `ax^3 + bx^2 + cx + d = 0`, with three roots, `alpha`, `beta`, and `gamma`, it is a rule that `alpha + beta + gamma = -b-:a`, also `alpha*beta + beta*gamma + gamma*alpha = c-:a`, and that `alpha*beta*gamma = -d-:a`.
Now because `d` is real and not imaginary, that means `alpha*beta*gamma` is real. You should know that to get a real number by multiplying imiginary numbers, you times it by it's conjugate, which for `2-3i` is `2+3i`. So now we know that `alpha` and `beta` are `2-3i` and `2+3i`.
We know `alpha*beta*gamma = -d-: a`, so `(2-3i)(2+3i)gamma = -(-39)-:1`. Expanding `(2-3i)(2+3i)` we get `13`, so `13gamma = 39`, thus `gamma = 39-:13 = 3`, therefore giving us the other two roots: `2+3i` and `3`.

Answer 2

`2+3i" and " 3`

Explanation:

`"the complex roots of polynomial equations always "`
`"occur in "color(blue)"conjugate pairs"`

`2-3i" is a root "rArr2+3i" is also a root"`

`"the quadratic factor formed by these roots is "`

`(z-(2-3i))(z-(2+3i))`

`=((z-2)+3i)((z-2)-3i)`

`=(z-2)^2-9i^2`

`=z^2-4z+4+9`

`=z^2-4z+13`

`rArrz^3-7z^2+25z-39`

`=color(red)(z)(z^2-4z+13)color(magenta)(+4z^2)-7z^2color(magenta)(-13z)+25z-39`

`=color(red)(z)(z^2-4z+13)color(red)(-3)(z^2-4z+13)color(magenta)(-12z)+12zcolor(magenta)(+39)`
`color(white)(=)-39`

`=color(red)(z)(z^2-4z+13)color(red)(-3)(z^2-4z+13)+0`

`rArr(z-3)" is a root"`

`rArr(z-3)(z-(2-3i))(z-(2+3i))=0`

`rArr"roots are " z=3" and " z=2+-3i`