How do you use the rational roots theorem to find all possible zeros of $f (x) = - 2 x^{3} + 3 x^{2} - 4 x + 8$ $f(x) = -2x^3 + 3x^2 -4x + 8$ ?

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How do you use the rational roots theorem to find all possible zeros of $f (x) = - 2 x^{3} + 3 x^{2} - 4 x + 8$ $f(x) = -2x^3 + 3x^2 -4x + 8$ ?

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Answer 1 · 2016-08-05T19:41:51

$f (x)$ $f(x)$ has no rational zeros.

It has one positive Real zero and a Complex conjugate pair of non-Real zeros.

Explanation:

$f (x) = - 2 x^{3} + 3 x^{2} - 4 x + 8$ $f(x) = -2x^3+3x^2-4x+8$

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 $8$ $8$ and $q$ $q$ a divisor of the coefficient $- 2$ $-2$ of the leading term.

That means that the only possible rational zeros are:

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

Note that $f (- x) = 2 x^{3} + 3 x^{2} + 4 x + 8$ $f(-x) = 2x^3+3x^2+4x+8$ has no changes of sign of coefficients. So by Descartes rule of signs, $f (x)$ $f(x)$ has no negative zeros.

So the only possible rational zeros are:

$\frac{1}{2}, 1, 2, 4, 8$ $1/2, 1, 2, 4, 8$

When $x = \pm 1$ $x=+-1$ all of the terms will be even except $3 x^{2} = 3$ $3x^2 = 3$ . Hence $f (\pm 1)$ $f(+-1)$ is odd and non-zero.

When $x = \pm 2$ $x=+-2$ all of the terms will be divisible by $8$ $8$ except $3 x^{2} = 12$ $3x^2 = 12$ . Hence $f (\pm 2) \neq 0$ $f(+-2) != 0$ .

When $x = \pm 4$ $x=+-4$ or $x = \pm 8$ $x=+-8$ all of the terms will be divisible by $16$ $16$ except the constant term $8$ $8$ . Hence $f (\pm 4) \neq 0$ $f(+-4) != 0$ and $f (\pm 8) \neq 0$ $f(+-8) != 0$ .

So that only leaves as a possible rational zero:

$\frac{1}{2}$ $1/2$

We find:

$f (\frac{1}{2}) = - 2 (\frac{1}{8}) + 3 (\frac{1}{4}) - 4 (\frac{1}{2}) + 8$ $f(1/2) = -2(1/8)+3(1/4)-4(1/2)+8$

$= \frac{- 1 + 3 - 8 + 32}{4}$ $= (-1+3-8+32)/4$

$= \frac{13}{2}$ $= 13/2$

So $f (x)$ $f(x)$ has no rational zeros.

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Bonus

What can we find out about the zeros of $f (x)$ $f(x)$ ?

The discriminant $Delta$ of a cubic polynomial in the form $ax^3+bx^2+cx+d$ is given by the formula:

$Delta = b^2c^2-4ac^3-4b^3d-27a^2d^2+18abcd$

In our example, $a=-2$ , $b=3$ , $c=-4$ , $d=8$ and we find:

$Delta = 144-512-864-6912+3456=-4688$

Since $Delta < 0$ , this cubic has exactly one Real zero (which we already know is positive) and two non-Real Complex zeros which are Complex conjugates of one another.

Let's simplify $f(x)$ by linear substitution (called a Tschirnhaus tranformation) as follows:

$0 = -4f(x) = 8x^3-12x^2+16x-32$

$=(2x-1)^3+5(2x-1)-26$

$=t^3+5t-26$

where $t = 2x-1$

Use Cardano's method to find the roots:

Let $t = u+v$

Then we want to solve:

$u^3+v^3+(3uv+5)(u+v)-26 = 0$

Add the constraint $v = -5/(3u)$ to eliminate the term in $(u+v)$ and get:

$u^3-125/(27u^3)-26 = 0$

Multiply through by $27u^3$ and rearrange to get:

$27(u^3)^2-702(u^3)-125 = 0$

Use the quadratic formula to find:

$u^3 = (702+-sqrt(702^2+4(27)(125)))/(2*27)$

$= (702+-sqrt(492804+13500))/54$

$= (702+-sqrt(506304))/54$

$= (702+-24sqrt(879))/54$

$= (351+-12sqrt(879))/27$

Since $u^3$ is Real and the derivation was symmetric in $u$ and $v$ , we can use one of these roots for $u^3$ and the other for $v^3$ to find the Real root:

$t_1 = 1/3(root(3)(351+12sqrt(879))+root(3)(351-12sqrt(879)))$

and related Complex roots:

$t_2 = 1/3(omega root(3)(351+12sqrt(879))+omega^2 root(3)(351-12sqrt(879)))$

$t_3 = 1/3(omega^2 root(3)(351+12sqrt(879))+omega root(3)(351-12sqrt(879)))$

where $omega = -1/2+sqrt(3)/2i$ is the primitive Complex cube root of $1$ .

Then $x = 1/2(1+t)$ hence $f(x)$ has Real zero:

$x_1 = 1/2(1+1/3(root(3)(351+12sqrt(879))+root(3)(351-12sqrt(879))))$

and related Complex zeros:

$x_2 = 1/2(1+1/3(omega root(3)(351+12sqrt(879))+omega^2 root(3)(351-12sqrt(879))))$

$x_3 = 1/2(1+1/3(omega^2 root(3)(351+12sqrt(879))+omega root(3)(351-12sqrt(879))))$

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How do you use the rational roots theorem to find all possible zeros of $f (x) = - 2 x^{3} + 3 x^{2} - 4 x + 8$ $f(x) = -2x^3 + 3x^2 -4x + 8$ ?

1 Answer

Explanation:

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How do you use the rational roots theorem to find all possible zeros of f(x) = -2x^3 + 3x^2 -4x + 8f(x)=−2x3+3x2−4x+8f(x) = -2x^3 + 3x^2 -4x + 8?

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How do you use the rational roots theorem to find all possible zeros of $f (x) = - 2 x^{3} + 3 x^{2} - 4 x + 8$ $f(x) = -2x^3 + 3x^2 -4x + 8$ ?