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

Jul 15, 2016

Zeros: $1 , - 1 , - \frac{1}{2} , \frac{2}{3} , 2$

#### Explanation:

Solve:

$6 {y}^{5} - 13 {y}^{4} - 6 {y}^{3} + 17 {y}^{2} - 4 = 0$

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 $\left(y - 1\right)$ a factor:

$6 {y}^{5} - 13 {y}^{4} - 6 {y}^{3} + 17 {y}^{2} - 4 = \left(y - 1\right) \left(6 {y}^{4} - 7 {y}^{3} - 13 {y}^{2} + 4 y + 4\right)$

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 $\left(y + 1\right)$ a factor:

$6 {y}^{4} - 7 {y}^{3} - 13 {y}^{2} + 4 y + 4 = \left(y + 1\right) \left(6 {y}^{3} - 13 {y}^{2} + 4\right)$

By the rational root theorem, any rational zeros of the remaining cubic must be expressible in the form $\frac{p}{q}$ 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:

$\pm \frac{1}{6} , \pm \frac{1}{3} , \pm \frac{1}{2} , \pm \frac{2}{3} , \pm 1 , \pm 2 , \pm 4$

Trying each in turn we find:

$f \left(- \frac{1}{2}\right) = - \frac{6}{8} - \frac{13}{4} + 4 = \frac{- 3 - 13 + 16}{4} = 0$

So $y = - \frac{1}{2}$ is a zero and $\left(2 y + 1\right)$ a factor:

$6 {y}^{3} - 13 {y}^{2} + 4 = \left(2 y + 1\right) \left(3 {y}^{2} - 8 y + 4\right)$

To factor the remaining quadratic, use an AC method:

Find a pair of factors of $A C = 3 \cdot 4 = 12$ with sum $B = 8$.

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

$3 {y}^{2} - 8 y + 4$

$= \left(3 {y}^{2} - 6 y\right) - \left(2 y - 4\right)$

$= 3 y \left(y - 2\right) - 2 \left(y - 2\right)$

$= \left(3 y - 2\right) \left(y - 2\right)$

So the remaining two zeros are: $y = \frac{2}{3}$ and $y = 2$.