# How do you find the number of possible positive real zeros and negative zeros then determine the rational zeros given f(x)=8x^3-6x^2-23x+6?

Sep 1, 2017

We can tell that $f \left(x\right)$ has $1$ negative real zero and $2$ or $0$ positive real zeros.

The zeros, which are all rational are: $2 , \frac{1}{4} , - \frac{3}{2}$

#### Explanation:

Given:

$f \left(x\right) = 8 {x}^{3} - 6 {x}^{2} - 23 x + 6$

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Descartes' Rule of Signs

The coefficients of $f \left(x\right)$ have signs in the pattern $+ - - +$. Since this has two changes (from $+$ to $-$ and from $-$ to $+$), Descartes' Rule of Signs tells us that $f \left(x\right)$ has $2$ or $0$ positive real zeros.

The coefficients of $f \left(- x\right)$ have signs in the pattern $- - + +$. With one change of signs, we can deduce that $f \left(x\right)$ has exactly one negative real zero.

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Rational root theorem

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 $6$ and $q$ a divisor of the coefficient $8$ of the leading term.

That means that the only possible rational zeros are:

$\pm \frac{1}{8} , \pm \frac{1}{4} , \pm \frac{3}{8} , \pm \frac{1}{2} , \pm \frac{3}{4} , \pm 1 , \pm \frac{3}{2} , \pm 2 , \pm 3 , \pm 6$

Note that the rational root theorem only tells us about possible rational zeros, not about all possible zeros which may be irrational or non-real complex.

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Actual zeros

It is probably easiest to try small integers first rather than fractions.

We find:

$f \left(1\right) = 8 - 6 - 23 + 6 = - 15$

$f \left(2\right) = 8 {\left(\textcolor{b l u e}{2}\right)}^{3} - 6 {\left(\textcolor{b l u e}{2}\right)}^{2} - 23 \left(\textcolor{b l u e}{2}\right) + 6 = 64 - 24 - 46 + 6 = 0$

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

$8 {x}^{3} - 6 {x}^{2} - 23 x + 6 = \left(x - 2\right) \left(8 {x}^{2} + 10 x - 3\right)$

We can factor the remaining quadratic using an AC method:

Find a pair of factors of $A C = 8 \cdot 3 = 24$ which differ by $B = 10$

The pair $12 , 2$ works.

Use this pair to split the middle term and factor by grouping:

$8 {x}^{2} + 10 x - 3 = \left(8 {x}^{2} + 12 x\right) - \left(2 x + 3\right)$

$\textcolor{w h i t e}{8 {x}^{2} + 10 x - 3} = 4 x \left(2 x + 3\right) - 1 \left(2 x + 3\right)$

$\textcolor{w h i t e}{8 {x}^{2} + 10 x - 3} = \left(4 x - 1\right) \left(2 x + 3\right)$

Hence the other two zeros are $x = \frac{1}{4}$ and $x = - \frac{3}{2}$

graph{(y-(8x^3-6x^2-23x+6))(60(x-2)^2+y^2-0.2)(60(x-1/4)^2+y^2-0.2)(60(x+3/2)^2+y^2-0.2) = 0 [-5, 5, -20, 20]}