How do you find all the zeros of #f(x) = x^3 + 13x^2 + 57x + 85#?

1 Answer
May 5, 2016

Answer:

#x = 5# or #x = -4+-i#

Explanation:

#f(x) = x^3+13x^2+57x+85#

By the rational roots 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 #85# and #q# a divisor of the coefficient #1# of the leading term.

That means that the only possible rational zeros are:

#+-1#, #+-5#, #+-17#, #+-85#

In addition, since all of the coefficients of #f(x)# are positive, it has no positive zeros, so that leaves:

#-1#, #-5#, #-17#, #-85#

Trying each of these in turn we find:

#f(-5) = -125+325-285+85 = 0#

So #x=-5# is a zero and #(x+5)# a factor:

#x^3+13x^2+57x+85 = (x+5)(x^2+8x+17)#

We can find the remaining two zeros by completing the square:

#0 = x^2+8x+17#

#=(x+4)^2-16+17#

#=(x+4)^2+1#

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

#=((x+4)-i)((x+4)+i)#

#=(x+4-i)(x+4+i)#

So #x = -4+-i#