# What are all the possible rational zeros for y=2x^5-x^4-26x^3+13x^2+72x-36 and how do you find all zeros?

Dec 8, 2016

The "possible" rational zeros are:

$\pm \frac{1}{2} , \pm 1 , \pm \frac{3}{2} , \pm 2 , \pm 3 , \pm 4 , \pm \frac{9}{2} , \pm 6 , \pm 9 , \pm 12 , \pm 18 , \pm 36$

The actual zeros are:

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

#### Explanation:

$f \left(x\right) = 2 {x}^{5} - {x}^{4} - 26 {x}^{3} + 13 {x}^{2} + 72 x - 36$

By the rational root theorem, any rational zeros of $f \left(x\right)$ are expressible in the form $\frac{p}{q}$ for integers $p , q$ with $p$ a divisor of the constant term $- 36$ and $q$ a divisor of the coefficient $2$ of the leading term.

That means that the only possible rational zeros are:

$\pm \frac{1}{2} , \pm 1 , \pm \frac{3}{2} , \pm 2 , \pm 3 , \pm 4 , \pm \frac{9}{2} , \pm 6 , \pm 9 , \pm 12 , \pm 18 , \pm 36$

In addition, note the the ratio of the first and second terms is identical to that of the third and fourth terms and the fifth and sixth terms. So this quintic will factor by grouping:

$2 {x}^{5} - {x}^{4} - 26 {x}^{3} + 13 {x}^{2} + 72 x - 36$

$= \left(2 {x}^{5} - {x}^{4}\right) - \left(26 {x}^{3} - 13 {x}^{2}\right) + \left(72 x - 36\right)$

$= {x}^{4} \left(2 x - 1\right) - 13 {x}^{2} \left(2 x - 1\right) + 36 \left(2 x - 1\right)$

$= \left({x}^{4} - 13 {x}^{2} + 36\right) \left(2 x - 1\right)$

We can factor the remaining quartic as a quadratic an ${x}^{2}$, then the resulting quadratic factors as difference of squares as follows:

$= \left({\left({x}^{2}\right)}^{2} - 13 \left({x}^{2}\right) + 36\right) \left(2 x - 1\right)$

$= \left({x}^{2} - 9\right) \left({x}^{2} - 4\right) \left(2 x - 1\right)$

$= \left({x}^{2} - {3}^{2}\right) \left({x}^{2} - {2}^{2}\right) \left(2 x - 1\right)$

$= \left(x - 3\right) \left(x + 3\right) \left(x - 2\right) \left(x + 2\right) \left(2 x - 1\right)$

So the zeros are:

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