How do you find all the zeros of #f(x)=x^4+6x^3+14x^2+54x+45#?

1 Answer
George C.
Aug 8, 2016

#f(x)# has zeros #-1, -5, +-3i#

Explanation:

#f(x) = x^4+6x^3+14x^2+54x+45#

First note that:

#f(-1) = 1-6+14-54+45 = 0#

So #x=-1# is a zero and #(x+1)# a factor:

#x^4+6x^3+14x^2+54x+45 = (x+1)(x^3+5x^2+9x+45)#

Looking at the remaining cubic, note that the ratio between the first and second terms is the same as that between the third and fourth terms. So this cubic will factor by grouping...

#x^3+5x^2+9x+45#

#=(x^3+5x^2)+(9x+45)#

#=x^2(x+5)+9(x+5)#

#=(x^2+9)(x+5)#

Hence zeros #x=-5# and #x=+-sqrt(-9)=+-3i#

Related questions
See all questions in Zeros
Impact of this question
2873 views around the world
You can reuse this answer
Creative Commons License