How do you find all zeros with multiplicities of f(x)=x5−2x4−4x+8?
1 Answer
May 29, 2017
The zeros are
Explanation:
Given:
f(x)=x5−2x4−4x+8
Note that the ratio between the first and second terms is the same as that between the third and fourth terms, so we can factor this quadrinomial by grouping:
x5−2x4−4x+8
=(x5−2x4)−(4x−8)
=x4(x−2)−4(x−2)
=(x4−4)(x−2)
=((x2)2−22)(x−2)
=(x2−2)(x2+2)(x−2)
=(x2−(√2)2)(x2−(√2i)2)(x−2)
=(x−√2)(x+√2)(x−√2i)(x+√2i)(x−2)
Hence zeros:
±√2 ,±√2i ,2
