How do you find all the zeros of #f(x)=x^3-2x^2-11x+52#?

1 Answer
Aug 4, 2016

Answer:

#f(x)# has zeros #-4# and #3+-2i#

Explanation:

#f(x) = x^3-2x^2-11x+52#

By the rational root theorem, any rational zeros of #f(x)# are expressible in the form #p/q# for integers #p, q# with #p# a divisor of the constant term #52# and #q# a divisor of the coefficient #1# of the leading term.

That means that the only possible rational zeros are:

#+-1, +-2, +-4, +-13, +-26, +-52#

In addition, note that the signs of the coefficients are in the pattern #+ - - +#, having #2# changes of sign, so by Descartes' rule of signs, #f(x)# has #2# or #0# positive Real zeros.

The pattern of signs of coefficients of #f(-x)# is #- - + +#. With just one change of sign, #f(x)# must have exactly one negative Real zero.

So check for negative rational zeros first:

#f(-1) = -1-2+11+52 = 60#

#f(-2) = -8-8+22+52 = 58#

#f(-4) = -64-32+44+52 = 0#

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

#x^3-2x^2-11x+52 = (x+4)(x^2-6x+13)#

The discriminant of the remaining quadratic factor is negative, so it has no Real zeros, but we can factor it with Complex coefficients:

#x^2-6x+13#

#=(x-3)^2-9+13#

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

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

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

Hence zeros: #x=3+-2i#