What are all the possible rational zeros for #f(x)=6x^3-x^2-13x+8# and how do you find all zeros?

1 Answer
Dec 24, 2016

Answer:

This cubic has "possible" rational zeros:

#+-1/6, +-1/3, +-2/3, +-1, +-4/3, +-2, +-8/3, +-4, +-8#

and actual zeros:

#1" "# and #" "-5/12+-sqrt(217)/12#

Explanation:

#f(x) = 6x^3-x^2-13x+8#

By the rational roots theorem, any rational zero of #f(x)# is expressible in the form #p/q# for integers #p, q# with #p# a divisor of the constant term #8# and #q# a divisor of the coefficient #6# of the leading term.

That means that the only possible rational zeros are:

#+-1/6, +-1/3, +-2/3, +-1, +-4/3, +-2, +-8/3, +-4, +-8#

There is a shortcut in that the sum of the coefficients is #0#.

That is:

#6-1-13+8 = 0#

So #f(1) = 0# and #(x-1)# is a factor:

#6x^3-x^2-13x+8 = (x-1)(6x^2+5x-8)#

The remaining quadratic factor is in the form #ax^2+bx+c# with #a=6#, #b=5# and #c=-8#.

This has discriminant #Delta# given by the formula:

#Delta = b^2-4ac = 5^2-4(6)(-8) = 25+192 = 217#

Since #Delta > 0# this quadratic has two Real zeros, but #Delta# is not a perfect square, so those zeros are irrational.

We can find the zeros using the quadratic formula:

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

#color(white)(x) = (-b+-sqrt(Delta))/(2a)#

#color(white)(x) = (-5+-sqrt(217))/12#

#color(white)(x) = -5/12+-sqrt(217)/12#