How do you find all roots for #x^3 - x^2 - 14x + 24 = 0#?

1 Answer
Jun 27, 2016

Answer:

#x=2#, #x=-4# and #x=3#

Explanation:

#f(x) = x^3-x^2-14x+24#

By the rational root 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 #24# 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, +-24#

Trying each in turn, we find:

#f(1) = 1-1-14+24 = 10#

#f(-1) = -1-1+14+24 = 36#

#f(2) = 8-4-28+24 = 0#

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

#x^3-x^2-14x+24 = (x-2)(x^2+x-12)#

We can factor the remaining quadratic by finding a pair of factors of the absolute value #12# of the constant term which differ by the coefficient #1# of the middle term. The pair #4,3# works.

Hence we find:

#x^2+x-12 = (x+4)(x-3)#

Putting it together:

#x^3-x^2-14x+24 = (x-2)(x+4)(x-3)#

which has zeros #x=2#, #x=-4# and #x=3#