How do you find all the real and complex roots of #F(x) = -9x^4 + 12x^3+40x^2-60x+25#?

1 Answer
May 26, 2016

Answer:

#x_1 = 2/3+1/3i#

#x_2 = 2/3-1/3i#

#x_3 = sqrt(5)#

#x_4 = -sqrt(5)#

Explanation:

Given:

#f(x) = -9x^4+12x^3+40x^2-60x+25#

By the rational root theorem, any rational zeros of #f(x)# will be expressible in the form #p/q# with integers #p, q#, with #p# a divisor of the constant term #25# and #q# a divisor of the coefficient #-9# of the leading term.

That means that the only possible rational zeros of #f(x)# are:

#+-1/9, +-1/3, +-5/9, +-1, +-5/3, +-25/9, +-5, +-25/3, +-25#

None of these work, so #f(x)# has no rational zeros.

Take a deep breath and start trying a full blown quartic solution:

Multiply through by #-9# to make the sums easier:

#-9f(x) = 81x^4-108x^3+360x^2-540x+225#

#=(3x-1)^4-46(3x-1)^2+92(3x-1)-88#

Let #t=3x-1#

We want to solve:

#0 = t^4-46t^2+92t-88#

#=(t^2-at+b)(t^2+at+c)#

#=t^4+(b+c-a^2)t^2+a(b-c)t+bc#

Equating coefficients and rearranging a little:

#{ (b+c = a^2-46), (b-c = 92/a), (bc = -88) :}#

Then:

#(a^2-46)^2 = (b+c)^2 = (b-c)^2+4bc = (92/a)^2-352#

Expanding both ends:

#a^4-92a^2+2116 = 8464/a^2-352#

Multiplying through by #a^2# and rearranging a little, this becomes:

#(a^2)^3-92(a^2)^2+2468(a^2)-8464 = 0#

Let #y=a^2#

Let #g(y) = y^3-92y^2+2468y-8464#

By the rational root theorem, any rational zeros of #g(y)# are factors of #8464#.

Trying the first few we find:

#g(4) = 64-1472+9872-8464=0#

So one solution is #y = 4#, that is #a^2=4#. So we can put #a=2#.

#{ (b+c = a^2-46 = 2^2-46 = -42), (b-c = 92/a = 92/2 = 46), (bc = -88) :}#

Hence:

#b=2# and #c=-44#

So we have:

#t^4-46t^2+92t-88 = (t^2-2t+2)(t^2+2t-44)#

which has zeros:

#t_1 = 1+i#

#t_2 = 1-i#

#t_3 = -1+3sqrt(5)#

#t_4 = -1-3sqrt(5)#

Then #x = (t+1)/3#, hence zeros of the original quartic:

#x_1 = 2/3+1/3i#

#x_2 = 2/3-1/3i#

#x_3 = sqrt(5)#

#x_4 = -sqrt(5)#