How do you find all the zeros of #f(x)=x^3-4x^2+16x-64#?
1 Answer
Aug 14, 2016
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#
