Question

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

Answer 1

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

Explanation:

`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`