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

##### 1 Answer

Jun 5, 2016

Roots

#### Explanation:

Note that the sum of the coefficients is

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

So

#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

#2+9-17+6 = 0#

So

#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

The pair

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