How do you find all the real solutions of the polynomial equation #2y^4+7y^3 -26y^2 +23y-6 =0#?

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Answer 1 · 2016-06-05T21:48:29

Roots #y=1# (multiplicity #2#), #y = 1/2# and #y=-6#

#2y^4+7y^3-26y^2+23y-6#

Note that the sum of the coefficients is #0#. That is:

#2+7-26+23-6 = 0#

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

#2y^4+7y^3-26y^2+23y-6 = (y-1)(2y^3+9y^2-17y+6)#

The sum of the coefficients of the remaining cubic factor also sum to #0#. That is:

#2+9-17+6 = 0#

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

#2y^3+9y^2-17y+6 = (y-1)(2y^2+11y-6)#

We can factor the remaining quadratic using an AC method:

Find a pair of factors of #AC = 2*6 = 12# which differ by #B=11#.

The pair #12, 1# works.

Use that pair to split the middle term, then factor by grouping:

#2y^2+11y-6#

#=2y^2+12y-y-6#

#=(2y^2+12y)-(y+6)#

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

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

So the remaining zeros are #y=1/2# and #y=-6#