# How do you find all roots for #6y^5 - 13y^4 - 6y^3 + 17y^2 - 4 = 0#?

##### 1 Answer

#### Answer:

Zeros:

#### Explanation:

Solve:

#6y^5-13y^4-6y^3+17y^2-4=0#

First note that the sum of the coefficients is

#6-13-6+17-4 = 0#

Hence

#6y^5-13y^4-6y^3+17y^2-4=(y-1)(6y^4-7y^3-13y^2+4y+4)#

Next note that if we reverse the signs of the coefficients of the terms with odd degree in the remaining quartic, then the sum is

#6+7-13-4+4 = 0#

Hence

#6y^4-7y^3-13y^2+4y+4 = (y+1)(6y^3-13y^2+4)#

By the rational root theorem, any *rational* zeros of the remaining cubic must be expressible in the form

That means that the only possible rational zeros are:

#+-1/6, +-1/3, +-1/2, +-2/3, +-1, +-2, +-4#

Trying each in turn we find:

#f(-1/2) = -6/8-13/4+4 = (-3-13+16)/4 = 0#

So

#6y^3-13y^2+4 = (2y+1)(3y^2-8y+4)#

To factor the remaining quadratic, use an AC method:

Find a pair of factors of

The pair

#3y^2-8y+4#

#= (3y^2-6y)-(2y-4)#

#= 3y(y-2)-2(y-2)#

#= (3y-2)(y-2)#

So the remaining two zeros are: