How do you use the rational root theorem to find the roots of #f(x)=x^4-x-4#?

1 Answer
Oct 24, 2015

Answer:

You can't. #f(x) = 0# has no rational roots.

Explanation:

By the rational root theorem, any rational root of #x^4-x-4 = 0# is expressible in lowest terms as #p/q# where #p, q in ZZ#, #q != 0#, #p# a divisor of the constant term #4# and #q# a divisor of the coefficient #1# of the leading term.

So the only possible rational roots are:

#+-1#, #+-2#, #+-4#

Then we find:

#f(-4) = 256+4-4 = 256#
#f(-2) = 16+2-4 = 14#
#f(-1) = 1+1-4 = -2#
#f(1) = 1-1-4 = -4#
#f(2) = 16-2-4 = 10#
#f(4) = 256-4-4 = 248#

So #f(x) = 0# has no rational roots.

Since #f(x)# is continuous, we can tell that #f(x) = 0# has Real zeros in #(-2, -1)# and #(1, 2)#.

Newton's method can then be used to find approximations for the two Real roots. Starting with an approximation #a_0#, find better approximations by iterating using the formula:

#a_(i+1) = a_i - f(x)/(f'(x))#

#f'(x) = 4x^3-1#

Starting with #a_0 = -1.5#, we find:

#a_1 = -1.3232758621#
#a_2 = -1.2853460646#
#a_3 = -1.2837842190#
#a_4 = -1.2837816659#
#a_5 = -1.2837816659#

Starting with #a_0 = 1.5#, we find:

#a_1 = 1.535#
#a_2 = 1.5337528051#
#a_3 = 1.5337511688#
#a_4 = 1.5337511688#

graph{x^4-x-4 [-10, 10, -5, 5]}