What are the possible rational roots of #x^4-5x^3+9x^2-7x+2=0# and then determine the rational roots?

1 Answer
Feb 4, 2017

Answer:

The "possible" rational roots are: #+-1#, #+-2#

The actual roots are: #1#, #1#, #1#, #2#.

Explanation:

Given:

#f(x) = x^4-5x^3+9x^2-7x+2#

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 #2# and #q# a divisor of the coefficient #1# of the leading term.

That means that the only possible rational zeros are:

#+-1, +-2#

Note also that the pattern of signs of the coefficients is: #+ - + - +#. With #4# changes, we can use Descartes' Rule of Signs to deduce that means that #f(x)# has #4#, #2# or #0# positive Real zeros. The signs of the coefficients of #f(-x)# are #+ + + + +#, meaning that #f(x)# has no negative zeros.

So the only possible rational zeros are #1# or #2#.

We find:

#f(1) = 1-5+9-7+2 = 0#

So #x=1# is a zero and #(x-1)# a factor:

#x^4-5x^3+9x^2-7x+2 = (x-1)(x^3-4x^2+5x-2)#

Note that the sum of the coefficients of the remaining cubic factor is also zero, so #x=1# is a zero again and #(x-1)# a factor:

#x^3-4x^2+5x-2 = (x-1)(x^2-3x+2)#

Note that the remaining quadratic also has #x=1# as a zero and #(x-1)# as a factor:

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