How do you use the rational roots theorem to find all possible zeros of #F(x) = 6x^3 - 20x^2 + 11x + 7#?

1 Answer
Jul 30, 2016

Answer:

Zeros: #7/3#, #1/2 +-sqrt(3)/2#

Explanation:

#F(x) = 6x^3-20x^2+11x+7#

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 #7# 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, +-1/2, +-1, +-7/6, +-7/3, +-7/2, +-7#

Trying each in turn, we (eventually) find:

#F(7/3) = 6(343/27)-20(49/9)+11(7/3)+7#

#=(686-980+231+63)/9 = 0#

So #x=7/3# is a zero and #(3x-7)# a factor:

#6x^3-20x^2+11x+7#

#=(3x-7)(2x^2-2x-1)#

Use the quadratic formula to find the zeros of the remaining quadratic:

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

#=(2+-sqrt(4+8))/4#

#=(2+-2sqrt(3))/4#

#=1/2+-sqrt(3)/2#