# How do you use the Rational Zeros theorem to make a list of all possible rational zeros, and use the Descarte's rule of signs to list the possible positive/negative zeros of f(x)=36x^4-12x^3-11x^2+2x+1?

Mar 17, 2018

$f \left(x\right)$ has zeros $\frac{1}{2}$ and $- \frac{1}{3}$, each with multiplicity $2$.

#### Explanation:

Given:

$f \left(x\right) = 36 {x}^{4} - 12 {x}^{3} - 11 {x}^{2} + 2 x + 1$

By the rational zeros 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 $1$ and $q$ a divisor of the coefficient $36$ of the leading term.

That means that the only possible rational zeros are:

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

Note that the pattern of signs of the coefficients of $f \left(x\right)$ is $+ - - + +$. With $2$ changes of sign, Descartes' Rule of Signs tells us that $f \left(x\right)$ has $2$ or $0$ positive real zeros.

The pattern of signs of the coefficients of $f \left(- x\right)$ is $+ + - - +$. With $2$ changes of sign, Descartes' Rule of Signs tells us that $f \left(x\right)$ has $2$ or $0$ negative real zeros.

Note that if we reverse the order of the coefficients, we get a polynomial whose zeros are the reciprocals of the zeros of $f \left(x\right)$:

$g \left(x\right) = {x}^{4} + 2 {x}^{3} - 11 {x}^{2} - 12 x + 36$

This has possible rational zeros:

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

(I prefer doing arithmetic with whole numbers)

We find:

$g \left(2\right) = {\left(2\right)}^{4} + 2 {\left(2\right)}^{3} - 11 \left({2}^{2}\right) - 12 \left(2\right) + 36 = 16 + 16 - 44 - 24 + 36 = 0$

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

${x}^{4} + 2 {x}^{3} - 11 {x}^{2} - 12 x + 36 = \left(x - 2\right) \left({x}^{3} + 4 {x}^{2} - 3 x - 18\right)$

We find that $2$ is also a zero of the remaining cubic:

${\left(2\right)}^{3} + 4 {\left(2\right)}^{2} - 3 \left(2\right) - 18 = 8 + 16 - 6 - 18 = 0$

So $\left(x - 2\right)$ is a factor again:

${x}^{3} + 4 {x}^{2} - 3 x - 18 = \left(x - 2\right) \left({x}^{2} + 6 x + 9\right)$

We can recognise the remaining quadratic factor as a perfect square trinomial:

${x}^{2} + 6 x + 9 = {\left(x + 3\right)}^{2}$

So the last two zeros of $g \left(x\right)$ are $x = - 3$ (with multiplicity $2$)

So $g \left(x\right) = {\left(x - 2\right)}^{2} {\left(x + 3\right)}^{2}$

and $f \left(x\right) = {\left(2 x - 1\right)}^{2} {\left(3 x + 1\right)}^{2}$ has zeros $\frac{1}{2} , \frac{1}{2} , - \frac{1}{3} , - \frac{1}{3}$