How do you find the number of possible positive real zeros and negative zeros then determine the rational zeros given #f(x)=x^4+2x^3-9x^2-2x+8#?

1 Answer
Dec 19, 2016

Answer:

The zeros of #f(x)# are:

#1, -1, -4, 2#

Explanation:

Given:

#f(x) = x^4+2x^3-9x^2-2x+8#

Note that the pattern of the signs of the coefficients is #+ + - - +#. By Descartes' Rule of Signs, since this has #2# changes of sign, #f(x)# has #2# or #0# positive Real zeros.

The pattern of the signs of #f(-x)# is #+ - - + +#. Since this has #2# changes of sign, #f(x)# has #2# or #0# negative Real zeros.

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 #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#

In addition, note that the sum of the coefficients of #f(x)# is #0#. That is:

#1+2-9-2-8 = 0#

Hence #f(1) = 0#, #x=1# is a zero and #(x-1)# a factor:

#x^4+2x^3-9x^2-2x+8 = (x-1)(x^3+3x^2-6x-8)#

Note that #x=-1# is a zero of the remaining cubic, since:

#-1+3+6-8 = 0#

So #(x+1)# is a factor:

#x^3+3x^2-6x-8 = (x+1)(x^2+2x-8)#

Finally, note that #4-2 = 2# and #4*2 = 8#. So we find:

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

So the zeros of #f(x)# are:

#1, -1, -4, 2#