How do you find all the zeros of #x^3-4x^2-44x+96#?

1 Answer
Aug 14, 2016

Answer:

This cubic has zeros: #2#, #8# and #-6#

Explanation:

#f(x) = x^3-4x^2-44x+96#

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

That means that the only possible rational zeros are:

#+-1, +-2, +-3, +-4, +-6, +-8, +-12, +-16, +-24, +-32, +-48, +-96#

Trying each in turn, we find:

#f(2) = 8-16-88+96 = 0#

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

#x^3-4x^2-44x+96 = (x-2)(x^2-2x-48)#

To factor the remaining quadratic find a pair of factors of #48# which differ by #2#. The pair #8, 6# works. Hence:

#x^2-2x-48 = (x-8)(x+6)#

Hence zeros: #8# and #-6#