How do you use the Rational Zeros theorem to make a list of all possible rational zeros, and use the Descarte's rule of signs to list the possible positive/negative zeros of #f(x)=36x^4-12x^3-11x^2+2x+1#?

1 Answer
Mar 17, 2018

Answer:

#f(x)# has zeros #1/2# and #-1/3#, each with multiplicity #2#.

Explanation:

Given:

#f(x) = 36x^4-12x^3-11x^2+2x+1#

By the rational zeros 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 #1# and #q# a divisor of the coefficient #36# of the leading term.

That means that the only possible rational zeros are:

#+-1/36, +-1/18, +-1/12, +-1/9, +-1/6, +-1/4, +-1/3, +-1/2, +-1#

Note that the pattern of signs of the coefficients of #f(x)# is #+ - - + +#. With #2# changes of sign, Descartes' Rule of Signs tells us that #f(x)# has #2# or #0# positive real zeros.

The pattern of signs of the coefficients of #f(-x)# is #+ + - - +#. With #2# changes of sign, Descartes' Rule of Signs tells us that #f(x)# has #2# or #0# negative real zeros.

Note that if we reverse the order of the coefficients, we get a polynomial whose zeros are the reciprocals of the zeros of #f(x)#:

#g(x) = x^4+2x^3-11x^2-12x+36#

This has possible rational zeros:

#+-1, +-2, +-3, +-4, +-6, +-9, +-12, +-18, +-36#

(I prefer doing arithmetic with whole numbers)

We find:

#g(2) = (2)^4+2(2)^3-11(2^2)-12(2)+36 = 16+16-44-24+36 =0#

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

#x^4+2x^3-11x^2-12x+36 = (x-2)(x^3+4x^2-3x-18)#

We find that #2# is also a zero of the remaining cubic:

#(2)^3+4(2)^2-3(2)-18 = 8+16-6-18 = 0#

So #(x-2)# is a factor again:

#x^3+4x^2-3x-18 = (x-2)(x^2+6x+9)#

We can recognise the remaining quadratic factor as a perfect square trinomial:

#x^2+6x+9 = (x+3)^2#

So the last two zeros of #g(x)# are #x=-3# (with multiplicity #2#)

So #g(x) = (x-2)^2(x+3)^2#

and #f(x) = (2x-1)^2(3x+1)^2# has zeros #1/2, 1/2, -1/3, -1/3#