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

1 Answer
Jul 15, 2016

Answer:

Zeros: #1, -1, -1/2, 2/3, 2#

Explanation:

Solve:

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

First note that the sum of the coefficients is #0#. That is:

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

Hence #y=1# is a zero and #(y-1)# a factor:

#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 #0#. That is:

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

Hence #y=-1# is a zero and #(y+1)# a factor:

#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 #p/q# for integers #p, q# with #p# a divisor of the constant term #4# and #q# a divisor of the coefficient #6# of the leading term.

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 #y=-1/2# is a zero and #(2y+1)# a factor:

#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 #AC=3*4=12# with sum #B=8#.

The pair #6, 2# works. Use this pair to split the middle term and factor by grouping:

#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: #y=2/3# and #y=2#.