How do you find all the zeros of #f(x)=x^3-4x^2+16x-64#?

1 Answer
Aug 14, 2016

Answer:

#f(x)# has zeros #4# and #+-4i#

Explanation:

Note that the ratio of the first and second terms is the same as that of the third and fourth terms. So this cubic factors by grouping:

#x^3-4x^2+16x-64#

#=(x^3-4x^2)+(16x-64)#

#=x^2(x-4)+16(x-4)#

#=(x^2+16)(x-4)#

#=(x^2-(4i)^2)(x-4)#

#=(x-4i)(x+4i)(x-4)#

Hence zeros:

#+-4i# and #4#