What are the possible rational zeros of #P(x) = 4x^4+19x^3-x^2+19x-5# ?

Question

Benjamin Achtenberg · Accepted Answer

The "possible" rational zeros are: #+-1/4, +-1/2, +-1, +-5/4, +-5/2, +-5#

The actual rational zeros are: #1/4# and #-5#

The other two zeros are: #+-i# Given: #P(x) = 4x^4+19x^3-x^2+19x-5# By the rational roots theorem, any rational zeros of #P(x)# are expressible in the form #p/q# for integers #p, q# with #p# a divisor of the constant term #-5# and #q# a divisor of the coefficient #4# of the leading term. That means that the only possible rational zeros are: #+-1/4, +-1/2, +-1, +-5/4, +-5/2, +-5# We find: #P(1/4) = 4(color(blue)(1/4))^4+19(color(blue)(1/4))^3-(color(blue)(1/4))^2+19(color(blue)(1/4))-5# #= 1/64+19/64-4/64+304/64-320/64 = 0# So #x=1/4# is a zero and #(4x-1)# a factor: #4x^4+19x^3-x^2+19x-5 = (4x-1)(x^3+5x^2+x+5)# #color(white)(4x^4+19x^3-x^2+19x-5) = (4x-1)((x^3+5x^2)+(x+5))# #color(white)(4x^4+19x^3-x^2+19x-5) = (4x-1)(x^2(x+5)+1(x+5))# #color(white)(4x^4+19x^3-x^2+19x-5) = (4x-1)(x^2+1)(x+5)# #color(white)(4x^4+19x^3-x^2+19x-5) = (4x-1)(x^2-i^2)(x+5)# #color(white)(4x^4+19x^3-x^2+19x-5) = (4x-1)(x-i)(x+i)(x+5)# So the rational zeros are #1/4# and #-5#. The other two zeros are #+-i#.