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

##### 1 Answer

#### Answer:

The rational roots are

The remaining two roots are

#### Explanation:

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

By the rational root theorem, any *rational* zeros of

That means that the only possible *rational* zeros are:

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

In addition, note that

So the only possible *rational* zeros of

#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

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

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

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

The last two zeros are

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