What are all the possible rational zeros for #f(x)=3x^3-8x^2-2x-3# and how do you find all zeros?

1 Answer
Oct 25, 2016

Answer:

#f(x)# has zeros #3# and #-1/6+-sqrt(11)/6i#

Explanation:

#f(x) = 3x^3-8x^2-2x-3#

By the rational roots 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 #-3# and #q# a divisor of the coefficient #3# of the leading term.

That means that the only possible rational zeros are:

#+-1/3, +-1, +-3#

In addition note that the signs of the coefficients of #f(x)# are in the pattern #+ - - -#. By Descartes' Rule of signs, since there is one change of sign, #f(x)# has exactly one positive Real zero.

So check the possible positive rational zeros first:

#f(1/3) = 3(1/27)-8(1/9)-2(1/3)-3 = (1-8-6-27)/9= -40/9#

#f(1) = 3-8-2-3 = -10#

#f(3) = 3(27)-8(9)-2(3)-3 = 81-72-6-3 = 0#

So #x=3# is a zero and #(x-3)# a factor:

#3x^3-8x^2-2x-3 = (x-3)(3x^2+x+1)#

Use the quadratic formula to solve #3x^2+x+1=0# to find the two remaining zeros:

#x = (-b+-sqrt(b^2-4ac))/(2a)#

#color(white)(x) = (-1+-sqrt(1^2-4(3)(1)))/(2*3)#

#color(white)(x) = (-1+-sqrt(1-12))/6#

#color(white)(x) = (-1+-sqrt(-11))/6#

#color(white)(x) = -1/6+-sqrt(11)/6i#