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
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
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
That is:
#6-1-13+8 = 0#
So
#6x^3-x^2-13x+8 = (x-1)(6x^2+5x-8)#
The remaining quadratic factor is in the form
This has discriminant
#Delta = b^2-4ac = 5^2-4(6)(-8) = 25+192 = 217#
Since
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#