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

Jun 5, 2016

Roots $y = 1$ (multiplicity $2$), $y = \frac{1}{2}$ and $y = - 6$

#### Explanation:

$2 {y}^{4} + 7 {y}^{3} - 26 {y}^{2} + 23 y - 6$

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

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

So $y = 1$ is a zero and $\left(y - 1\right)$ a factor:

$2 {y}^{4} + 7 {y}^{3} - 26 {y}^{2} + 23 y - 6 = \left(y - 1\right) \left(2 {y}^{3} + 9 {y}^{2} - 17 y + 6\right)$

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

$2 {y}^{3} + 9 {y}^{2} - 17 y + 6 = \left(y - 1\right) \left(2 {y}^{2} + 11 y - 6\right)$

We can factor the remaining quadratic using an AC method:

Find a pair of factors of $A C = 2 \cdot 6 = 12$ which differ by $B = 11$.

The pair $12 , 1$ works.

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

$2 {y}^{2} + 11 y - 6$

$= 2 {y}^{2} + 12 y - y - 6$

$= \left(2 {y}^{2} + 12 y\right) - \left(y + 6\right)$

$= 2 y \left(y + 6\right) - 1 \left(y + 6\right)$

$= \left(2 y - 1\right) \left(y + 6\right)$

So the remaining zeros are $y = \frac{1}{2}$ and $y = - 6$