How do you find all rational roots for #8y^4 - 6y^3 + 17y^2 - 12y + 2 = 0#?

1 Answer
Aug 4, 2016

Answer:

The rational roots are #1/4#, #1/2#.

The remaining two roots are #+-sqrt(2)i#

Explanation:

#f(y)=8y^4-6y^3+17y^2-12y+2#

By the rational root theorem, any rational zeros of #f(y)# must be 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 #8# of the leading term.

That means that the only possible rational zeros are:

#+-1/8, +-1/4, +-1/2, +-1, +-2#

In addition, note that #f(-y) = 8y^4+6y^3+17y^2+12y+2# has all positive coefficients. Hence #f(y)# has no negative zeros.

So the only possible rational zeros of #f(y)# are:

#1/8, 1/4, 1/2, 1, 2#

We find:

#f(1/4) = 8(1/4)^4-6(1/4)^3+17(1/4)^2-12(1/4)+2#

#=(1-3+34-96+64)/32 = 0#

#f(1/2) = 8(1/2)^4-6(1/2)^3+17(1/2)^2-12(1/2)+2#

#=(2-3+17-24+8)/4 = 0#

So #y=1/4# and #y=1/2# are zeros and #(4y-1)# and #(2y-1)# are factors:

#8y^4-6y^3+17y^2-12y+2#

#=(4y-1)(2y^3-y^2+4y-2)#

#=(4y-1)(2y-1)(y^2+2)#

#y^2+2 >= 2 > 0# for all Real values of #y#, so there are no more Real, let alone rational, zeros.

The last two zeros are #+-sqrt(2)i# since:

#(y-sqrt(2)i)(y+sqrt(2)i) = y^2-(sqrt(2)i)^2 = y^2+2#