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

Question

Lincoln Anderson · Accepted Answer

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

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

James Allen · Answer

undefined The rational root theorem states that any rational root of a polynomial equation $a_nx^n + a_{n-1}x^{n-1} + \ldots + a_1x + a_0 = 0$ must be of the form $ \pm \frac{p}{q} $, where $p$ is a factor of the constant term $a_0$ and $q$ is a factor of the leading coefficient $a_n$.

For the equation $8y^4 - 6y^3 + 17y^2 - 12y + 2 = 0$, the constant term is 2 and the leading coefficient is 8.

The factors of 2 are ±1, ±2, and the factors of 8 are ±1, ±2, ±4, ±8.

Therefore, the possible rational roots are:

±1, ±2, ±1/2, ±1/4