How do you use the rational roots theorem to find all possible zeros of #f(x) = x^3 - 12x - 16#?

1 Answer
Jun 18, 2016

Answer:

The zeros of this #f(x)# are all rational:

#x=-2# with multiplicity #2#

#x=4#

Explanation:

#f(x) = x^3-12x-16#

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 #-16# and #q# a divisor of the coefficient #1# of the leading term.

That means that the only possible rational zeros are:

#+-1#, #+-2#, #+-4#, #+-8#, #+-16#

Also note that the signs of the coefficients of #f(x)# follow the pattern #+ - -#, with one change of sign. Hence #f(x)# has exactly one positive zero.

So start trying the positive possibilities first:

#f(1) = 1-12-16 = -27#

#f(2) = 8-24-16 = -32#

#f(4) = 64-48-16 = 0#

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

#x^3-12x-16 = (x-4)(x^2+4x+4) = (x-4)(x+2)(x+2)#

So #x=-2# is the other zero of #f(x)#, with multiplicity #2#