# 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)=2x^4+x^3-7x^2-3x+3?

May 31, 2018

$2 {x}^{4} + {x}^{3} - 7 {x}^{2} - 3 x + 3 = \left(2 x - 1\right) \left(x + 1\right) \left({x}^{2} - 3\right)$

#### Explanation:

It iseasy to find
$x = - 1$
By the rational root theórem we get
$x = \frac{1}{2}$
Dividing

$2 {x}^{4} + {x}^{3} - 7 {x}^{2} - 3 x + 3$ by
$2 {x}^{2} + x - 1$ we get
${x}^{2} - 3$

Aug 10, 2018

See explanation...

#### Explanation:

Given:

$f \left(x\right) = 2 {x}^{4} + {x}^{3} - 7 {x}^{2} - 3 x + 3$

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 $3$ and $q$ a divisor of the coefficient $2$ of the leading term.

That means that the only possible rational zeros are:

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

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

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

Trying each of the possible rational zeros in turn we find:

$f \left(\textcolor{b l u e}{\frac{1}{2}}\right) = 2 {\left(\textcolor{b l u e}{\frac{1}{2}}\right)}^{4} + {\left(\textcolor{b l u e}{\frac{1}{2}}\right)}^{3} - 7 {\left(\textcolor{b l u e}{\frac{1}{2}}\right)}^{2} - 3 \left(\textcolor{b l u e}{\frac{1}{2}}\right) + 3$

$\textcolor{w h i t e}{f \left(\frac{1}{2}\right)} = \frac{2}{16} + \frac{1}{8} - \frac{7}{4} - \frac{3}{2} + 3$

$\textcolor{w h i t e}{f \left(\frac{1}{2}\right)} = 0$

So $x = \frac{1}{2}$ is a zero and $\left(2 x - 1\right)$ is a factor:

$2 {x}^{4} + {x}^{3} - 7 {x}^{2} - 3 x + 3 = \left(2 x - 1\right) \left({x}^{3} + {x}^{2} - 3 x - 3\right)$

$\textcolor{w h i t e}{2 {x}^{4} + {x}^{3} - 7 {x}^{2} - 3 x + 3} = \left(2 x - 1\right) \left({x}^{2} \left(x + 1\right) - 3 \left(x + 1\right)\right)$

$\textcolor{w h i t e}{2 {x}^{4} + {x}^{3} - 7 {x}^{2} - 3 x + 3} = \left(2 x - 1\right) \left({x}^{2} - 3\right) \left(x + 1\right)$

So the other zeros are $x = \pm \sqrt{3}$ and $x = - 1$

graph{2x^4+x^3-7x^2-3x+3 [-2.5, 2.5, -6, 5]}