What are all the possible rational zeros for #f(x)=x^3+5x^2-11x-6# and how do you find all zeros?

1 Answer
Oct 26, 2016

Answer:

#f(x)# has zeros #2# and #-7/2+-sqrt(37)/2#

Explanation:

#f(x) = x^3+5x^2-11x-6#

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

Thats means that the only possible rational zeros are:

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

In addition, note that the signs of the coefficients of #f(x)# are in the pattern: #+ + - -#. With one change of sign, Descartes' Rule of Signs tells us that #f(x)# has exactly one positive Real zero. So let's try the positive rational possibilities first:

#f(1) = 1+5-11+6 = 1#

#f(2) = 8+5(4)-11(2)-6 = 8+20-22-6 = 0#

So #x=2# is a zero and #(x-2)# a factor of #f(x)#:

#x^3+5x^2-11x-6 = (x-2)(x^2+7x+3)#

The remaining two zeros are roots of the quadratic: #x^2+7x+3 = 0#, which we can solve using the quadratic formula:

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

#color(white)(x) = (-7+-sqrt(7^2-4(1)(3)))/(2*1)#

#color(white)(x) = (-7+-sqrt(49-12))/2#

#color(white)(x) = -7/2+-sqrt(37)/2#