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.
