# 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)=x^3-2x^2-5x+6?

Jul 2, 2018

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#### Explanation:

Given:

$f \left(x\right) = {x}^{3} - 2 {x}^{2} - 5 x + 6$

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 $6$ and $q$ a divisor of the coefficient $1$ of the leading term.

That means that the only possible rational zeros are:

$\pm 1 , \pm 2 , \pm 3 , \pm 6$

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 $0$ or $2$ positive real zeros.

The pattern of signs of the coefficients of $f \left(- x\right)$ is $- - + +$. With one change of sign, Descartes' Rule of Signs tells us that $f \left(x\right)$ has exactly one negative real zero.

In addition, note that the sum of the coefficients of $f \left(x\right)$ is zero. That is:

$1 - 2 - 5 + 6 = 0$

Hence we can tell that $f \left(1\right) = 0$ and $\left(x - 1\right)$ is a factor of $f \left(x\right)$:

${x}^{3} - 2 {x}^{2} - 5 x + 6 = \left(x - 1\right) \left({x}^{2} - x - 6\right) = \left(x - 1\right) \left(x - 3\right) \left(x + 2\right)$

So the other two zeros of $f \left(x\right)$ are $x = 3$ and $x = - 2$