How do you use the rational root theorem to find the roots of #8y^4 - 6y^3 + 17y^2 - 12y + 2 = 0#?

1 Answer
Jul 8, 2015

Answer:

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

has two rational roots #1/2# and #1/4# findable by the rational roots theorem.

Explanation:

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

By the rational root theorem, any rational roots of #f(y) = 0# must be of the form #p/q# where #p# and #q# are integers, in lowest terms such that #p# is a divisor of the constant term #2# and #q# is a divisor of the coefficient, #8#, of the highest order term.

So the possible rational roots are:

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

Notice that #f(-x) = 8x^4+6x^3+17x^2+12x+2# has all positive coefficients, so no positive value of #x# is a root of #f(-x) = 0#. Hence no negative value of #y# is a root of #f(y) = 0#.

So the only possible rational roots are:

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

#f(2) = 8*16-6*8+17*4-12*2+2#

#= 128-48+68-24+2 = 126#

#f(1) = 8-6+17-12+2 = 9#

#f(1/2) = 8/16-6/8+17/4-12/2+2#

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

#f(1/4) = 8/256-6/64+17/16-12/4+2#

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

We can tell in advance that #f(1/8) != 0#, since we have already found roots #y=1/2# and #y=1/4#, which give factors #(2y-1)# and #(4y-1)#, which together will exhaust the leading coefficient of #y^4#.

#(2y-1)(4y-1) = 8y^2-6y+1#

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

#=(8y^2-6y+1)(y^2+2)#

Notice that #y^2+2 >= 2 > 0# for all #y in RR#, so there are no more real roots.