How do you find all the real and complex roots of #p(x) = x^5 + 4x^4 + 6x^3 + 4x^2 + 5x#?

1 Answer
Jan 11, 2016

Answer:

The zeros are #x=0#, #x=+-i#, #x=-2+-i#

Explanation:

#p(x) = x^5+4x^4+6x^3+4x^2+5x#

First separate out the common factor #x# to find:

#p(x) = x(x^4+4x^3+6x^2+4x+5)#

Note that #x=0# is one of the zeros of #p(x)#.

Next, let #t = x+1#

This is called a Tschirnhaus transformation. This kind of simple Tschirnhaus transformation can be used normally to eliminate the #x^3# term, but in the case of this particular quartic it helps a lot more.

Then #t^4 = x^4+4x^3+6x^2+4x+1#

So #x^4+4x^3+6x^2+4x+5 = t^4+4#

This has zeros when #t^4 = -4#, that is when #t^2 = +-sqrt(-4) = +-2i#

That is when #t = +-1+-i#

If #t = 1+-i# then #x = t - 1 = +-i#

If #t = -1+-i# then #x = t - 1 = -2+-i#