How do you find all the zeros of #f(x)=3x^4-17x^3+33x^2-17x-10#?

1 Answer
Apr 16, 2016

Answer:

Find zeros: #-1/3#, #2# and #2+-i#

Explanation:

By the rational root theorem, any rational zeros of #f(x)# must be expressible in the form #p/q# for integers #p, q# with #p# a divisor of the constant term #-10# and #q# a divisor of the coefficient #3# of the leading term.

That means that the only possible rational zeros are:

#+-1/3, +-2/3, +-1, +-5/3, +-2, +-10/3, +-5, +-10#

Trying each in turn, we find:

#f(-1/3) = 3/81+17/27+33/9+17/3-10#

#=(1+17+99+153-270)/27 = 0#

#f(2) = 48-136+132-34+10 = 0#

So #x=-1/3# and #x=2# are zeros and #(3x+1)# and #(x-2)# are factors of #f(x)#:

#3x^4-17x^3+33x^2-17x-10#

#=(3x+1)(x^3-6x^2+13x-10)#

#=(3x+1)(x-2)(x^2-4x+5)#

The remaining quadratic factor has negative discriminant, but we can factor it using Complex numbers, completeing the square and the difference of squares identity:

#a^2-b^2 = (a-b)(a+b)#

as follows:

#x^2-4x+5 = (x-2)^2+1 = (x-2)^2-i^2 = ((x-2)-i)((x-2)+i) = (x-2-i)(x-2+i)#

So the remaining zeros are #x = 2+-i#