How do you use the rational roots theorem to find all possible zeros of #P(x) = x^5 -2x^4 +4x -8#?

1 Answer
Apr 26, 2016

Answer:

The only Real zero is #x=2#

Explanation:

By the rational roots theorem, any rational zeros of #P(x)# will be expressible in the form #p/q# for integers #p, q# with #p# a divisor of the constant term #-8# and #q# a divisor of the coefficient #1# of the leading term.

That means that the only possible rational zeros are:

#+-1#, #+-2#, #+-4#, #+-8#

Trying each in turn, we find:

#P(2) = 32-32+8-8 = 0#

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

#x^5-2x^4+4x-8 = (x-2)(x^4+4)#

Note that #x^4 >= 0# for all Real values of #x# so #x^4+4 >= 4# has no Real zeros.

It does have Complex zeros #x=1+-i# and #x=-1+-i#, being the #4#th roots of #-4#.

#color(white)()#
Bonus

Note that although #x^4+4# has no Real zeros or linear factors, it is possible to factor it into a product of two quadratics with Real coefficients.

In fact, we find:

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

This is an instance of a nice identity for factoring quartics:

#(a^2-kab+b^2)(a^2+kab+b^2) = a^4+(2-k^2)a^2b^2+b^4#

Note in particular that if we put #k=sqrt(2)# we find:

#(a^2-sqrt(2)ab+b^2)(a^2+sqrt(2)ab+b^2) = a^4+b^4#

In our case we used this with #k=sqrt(2)#, #a=x# and #b=sqrt(2)#