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

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

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Answer 1 · 2017-09-01T08:29:13

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

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

Explanation:

Given:

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

$color(white)()$
Descartes' Rule of Signs

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

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

$color(white)()$
Rational root theorem

By the rational root theorem, any rational zeros of $f (x)$ $f(x)$ are expressible in the form $\frac{p}{q}$ $p/q$ for integers $p, q$ $p, q$ with $p$ $p$ a divisor of the constant term $6$ $6$ and $q$ $q$ a divisor of the coefficient $8$ $8$ of the leading term.

That means that the only possible rational zeros are:

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

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.

$color(white)()$
Actual zeros

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

We find:

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

$f (2) = 8 {(2)}^{3} - 6 {(2)}^{2} - 23 (2) + 6 = 64 - 24 - 46 + 6 = 0$ $f(2) = 8(color(blue)(2))^3-6(color(blue)(2))^2-23(color(blue)(2))+6 = 64-24-46+6 = 0$

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

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

We can factor the remaining quadratic using an AC method:

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

The pair $12, 2$ $12, 2$ works.

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

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

$8 x^{2} + 10 x - 3 = 4 x (2 x + 3) - 1 (2 x + 3)$ $color(white)(8x^2+10x-3) = 4x(2x+3)-1(2x+3)$

$8 x^{2} + 10 x - 3 = (4 x - 1) (2 x + 3)$ $color(white)(8x^2+10x-3) = (4x-1)(2x+3)$

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

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]}

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

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How do you find the number of possible positive real zeros and negative zeros then determine the rational zeros given f(x)=8x3−6x2−23x+6f(x)=8x^3-6x^2-23x+6?

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