Question

How do you factorize `-9z^2+24z-32` over C?

Answer 1

`-9z^2+24z-32=(x-4/3+sqrt(15)/3 i)(x-4/3-sqrt(15)/3 i)`

Explanation:

If by "factorize...over C", you mean the set of complex numbers, then here are the following steps:

Step 1. Set the equation equal to zero
`-9z^2+24z-32=0`

Step 2. Plug the coefficients into the quadratic equation
`z=(-b+-sqrt(b^2-4ac))/(2a)` where `a=-9`, `b=24`, and `c=-32`

This gives

`z=(-24+-sqrt(24^2-4(-9)(-31)))/(2(-9))`

Step 3. Simplify
`z=(-24+-sqrt(576-1116))/(-18)`
`z=(-24+-sqrt(-540))/-18`
`z=(-24+-sqrt(-36*15))/-18`
`z=(-24+-6sqrt(-15))/-18`
`z=(-24+-6sqrt(15)i)/-18`
`z=-24/-18+-(6sqrt(15)i)/-18`
`z=4/3+-(-sqrt(15))/3 i`

Step 3. Break into two parts

`z=4/3-sqrt(15)/3 i` and `z=4/3+sqrt(15)/3 i`

`z-4/3+sqrt(15)/3 i=0` and `z-4/3-sqrt(15)/3 i=0`

Step 4. Multiply these terms together for the final answer
`-9z^2+24z-32=(z-4/3+sqrt(15)/3 i)(z-4/3-sqrt(15)/3 i)`