# How do you find all rational roots for 4y^5 + 8y^4 - 29y^3 - 42y^2 + 45y + 54 = 0?

Aug 4, 2016

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

#### Explanation:

$f \left(y\right) = 4 {y}^{5} + 8 {y}^{4} - 29 {y}^{3} - 42 {y}^{2} + 45 y + 54$

By the rational root theorem, any rational zeros of $f \left(y\right)$ are expressible in the form $\frac{p}{q}$ for integers $p , q$ with $p$ a divisor of the constant term $54$ and $q$ a divisor of the coefficient $4$ of the leading term.

That means that the only possible rational zeros are:

$\pm \frac{1}{4} , \pm \frac{1}{2} , \pm \frac{3}{4} , \pm 1 , \pm \frac{3}{2} , \pm 2 , \pm \frac{9}{4} , \pm 3 , \pm \frac{9}{2} , \pm 6 , \pm \frac{27}{4} , \pm 9 , \pm \frac{27}{2} , \pm 18 , \pm 27 , \pm 54$

That's rather a lot of possibilities to try, but first note that:

$f \left(- 1\right) = - 4 + 8 + 29 - 42 - 45 + 54 = 0$

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

$4 {y}^{5} + 8 {y}^{4} - 29 {y}^{3} - 42 {y}^{2} + 45 y + 54$

$= \left(y + 1\right) \left(4 {y}^{4} + 4 {y}^{3} - 33 {y}^{2} - 9 y + 54\right)$

Sticking with simpler calculations, let's try substituting $y = 2$ in the remaining quartic:

$4 {y}^{4} + 4 {y}^{3} - 33 {y}^{2} - 9 y + 54$

$= 4 \left(16\right) + 4 \left(8\right) - 33 \left(4\right) - 9 \left(2\right) + 54$

$= 64 + 32 - 132 - 18 + 54 = 0$

So $y = 2$ is a zero and $\left(y - 2\right)$ a factor:

$4 {y}^{4} + 4 {y}^{3} - 33 {y}^{2} - 9 y + 54 = \left(y - 2\right) \left(4 {y}^{3} + 12 {y}^{2} - 9 y - 27\right)$

In the remaining cubic, the ratio of the first and second terms is the same as the ratio between the third and fourth terms. Hence it will factor by grouping:

$4 {y}^{3} + 12 {y}^{2} - 9 y - 27$

$= \left(4 {y}^{3} + 12 {y}^{2}\right) - \left(9 y + 27\right)$

$= 4 {y}^{2} \left(y + 3\right) - 9 \left(y + 3\right)$

$= \left(4 {y}^{2} - 9\right) \left(y + 3\right)$

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

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

Hence three more zeros: $\frac{3}{2}$, $- \frac{3}{2}$, $- 3$