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

1 Answer
May 22, 2017

Answer:

#-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)#