How do you find the roots of #x^3-x^2-34x-56=0#?

1 Answer
Nov 8, 2016

Answer:

This cubic equation has roots #7, -2, -4#.

Explanation:

#f(x) = x^3-x^2-34x-56#

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 #-56# and #q# a divisor of the coefficient #1# of the leading term.

That means that the only possible rational zeros are:

#+-1, +-2, +-4, +-7, +-8, +-14, +-28, +-56#

In addition, note that the pattern of signs of the coefficients of the terms is #+ - - -#. By Descartes' Rule of Signs, since this has only one change of sign, #f(x)# has exactly one positive Real zero.

If #x >= 1# then #x^2+34x+56 >= 1+34+56 = 91#. So that positive Real zero must satisfy #x^3 >= 91#. So we can immediately rule out #1, 2, 4# as possibilities. Try #x = 7#:

#f(7) = 7^3-7^2-34(7)-56 = 343-49-238-56 = 0#

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

#x^3-x^2-34x-56 = (x-7)(x^2+6x+8)#

We can factor the remaining quadratic by finding a pair of numbers with sum #6# and product #8#. The pair #2, 4# works, so we have:

#x^2+6x+8 = (x+2)(x+4)#

Putting it all together:

#x^3-x^2-34x-56 = (x-7)(x+2)(x+4)#

with zeros #7, -2, -4#