How do you use the rational root theorem to find the roots of #2x^4 + 3x^3 - 7x^2 + 3x - 9 = 0 #?

1 Answer
George C.
May 23, 2015

Any rational root #p/q# - if written in lowest terms, so that #p# and #q# have no common factor other than 1, satisfies the property that #p# is a divisor of the constant term #-9# and #q# is a divisor of the coefficient #2# of the highest order term (#2x^4#).

So the only possible rational roots are:

#+-1/2#, #+-1#, #+-3/2#, #+-3#, #+-9/2# or #+-9#.

Rustling up a quick spreadsheet to help, I found that #-3# and #3/2# are roots, hence #(x+3)# and #(2x-3)# are factors of #2x^4+3x^3-7x^2+3x-9#.

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

Use synthetic division to find:

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

So the other 2 roots of #2x^4+3x^3-7x^2+3x-9 = 0# are #+-i#

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