How do you find all the zeros of #f(x) = x^4 - 9x^3 + 24x^2 - 6x - 40#?

1 Answer
May 8, 2016

Answer:

#x = -1#, #x = 4#, #x = 3+-i#

Explanation:

First notice that if you reverse the signs of the coefficients on the terms of odd degree then the sum is #0#. That is:

#1+9+24+6-40 = 0#

So #f(-1) = 0# and #(x+1)# is a factor:

#x^4-9x^3+24x^2-6x-40 = (x+1)(x^3-10x^2+34x-40)#

Let #g(x) = x^3-10x^2+34x-40#

By the rational root theorem, any rational zeros of #g(x)# must be expressible in the form #p/q# for integers #p#, #q# with #p# a divisor of the constant term #-40# and #q# a divisor of the coefficient #1# of the leading term.

That means that the only possible rational zeros are:

#+-1#, #+-2#, #+-4#, #+-5#, #+-8#, #+-10#, #+-20#, #+-40#

In addition, note the #g(-x) = -x^3-10x^2-34x-40# has all negative coefficients. So there are no negative zeros.

So the only possible rational zeros of #g(x)# are:

#1, 2, 4, 5, 8, 10, 20, 40#

We find:

#g(4) = 64-160+136-40 = 0#

So #x=4# is a zero and #(x-4)# a factor:

#x^3-10x^2+34x-40#

#=(x-4)(x^2-6x+10)#

The remaining quadratic factor has negative discriminant, so its zeros are Complex, but we can factor it as a difference of squares:

#x^2-6x+10#

#=(x-3)^2-9+10#

#=(x-3)^2+1#

#=(x-3)^2-i^2#

#=((x-3)-i)((x-3)+i)#

#=(x-3-i)(x-3+i)#

So the remaining zeros are:

#x = 3+-i#