# How do you find all rational zeroes of the function using synthetic division f(x)=2x^5+x^4-23x-16?

Feb 2, 2018

$f \left(x\right)$ has three irrational real zeros and two non-real complex zeros.

#### Explanation:

Given:

$f \left(x\right) = 2 {x}^{5} + {x}^{4} - 23 x - 16$

By Descartes' Rule of Signs $f \left(x\right)$ has exactly one positive real zero, since the signs of its coefficients $+ + - -$ have only one change of signs.

$f \left(- x\right)$ has signs in the pattern $- + + -$. With $2$ changes of sign, we can deduce that $f \left(x\right)$ has $0$ or $2$ negative real zeros.

By the rational roots 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 $- 16$ and $q$ a divisor of the coefficient $2$ of the leading term.

That means that the only possible rational zeros are:

$\pm \frac{1}{2} , \pm 1 , \pm 2 , \pm 4 , \pm 8 , \pm 16$

We find:

$f \left(- 2\right) = - 64 + 16 + 46 - 16 = - 18 < 0$

$f \left(- 1\right) = - 2 + 1 + 23 - 16 = 6 > 0$

$f \left(- \frac{1}{2}\right) = - \frac{1}{16} + \frac{1}{16} + \frac{23}{2} - 16 = - \frac{9}{2} < 0$

$f \left(1\right) = 2 + 1 - 23 - 16 = - 36 < 0$

$f \left(2\right) = 64 + 16 - 46 - 16 = 18 > 0$

Hence $f \left(x\right)$ has irrational zeros in $\left(- 2 , - 1\right)$, $\left(- 1 , - \frac{1}{2}\right)$ and $\left(1 , 2\right)$

The remaining $2$ roots are non-real complex.

We cannot immediately deduce it from what we have found, but actually none of the zeros are expressible in terms of $n$th roots (i.e. square, cube roots etc.).