Question

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

Answer 1

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`.