How do you list all possible roots and find all factors of $x^{4} + 2 x^{3} - 8 x^{2} + 16 x - 23$ $x^4+2x^3-8x^2+16x-23$ ?

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How do you list all possible roots and find all factors of $x^{4} + 2 x^{3} - 8 x^{2} + 16 x - 23$ $x^4+2x^3-8x^2+16x-23$ ?

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Answer 1 · 2017-12-21T20:05:09

Here is a solution for $x^{4} - 2 x^{3} + 8 x^{2} + 16 x - 23 = 0$ $x^4-2x^3+8x^2+16x-23=0$

( $x^{4} + 2 x^{3} - 8 x^{2} + 16 x - 23 = 0$ $x^4+2x^3-8x^2+16x-23=0$ is much more complicated)

Explanation:

Assuming typographical errors in the problem, let us solve:

$x^{4} - 2 x^{3} + 8 x^{2} + 16 x - 23 = 0$ $x^4-2x^3+8x^2+16x-23=0$

By the rational roots theorem, any rational zeros of this quartic 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 $- 23$ $-23$ and $q$ $q$ a divisor of the coefficient $1$ $1$ of the leading term.

That means that the only possible rational zeros are:

$\pm 1, \pm 23$ $+-1, +-23$

Note that $x = 1$ $x=1$ is a solution, since the sum of the coefficients is $0$ $0$ , i.e. $1 - 2 + 8 + 16 - 23 = 0$ $1-2+8+16-23 = 0$

Then:

$x^{4} - 2 x^{3} + 8 x^{2} + 16 x - 23 = (x - 1) (x^{3} - x^{2} + 7 x + 23)$ $x^4-2x^3+8x^2+16x-23= (x-1)(x^3-x^2+7x+23)$

To find the zeros for the remaining cubic, proceed as follows:

$f (x) = x^{3} - x^{2} + 7 x + 23$ $f(x) = x^3-x^2+7x+23$

Discriminant

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, $a=1$ , $b=-1$ , $c=7$ and $d=23$ , so we find:

$Delta = 49-1372+92-14283-2898 = -18412$

Since $Delta < 0$ this cubic has $1$ Real zero and $2$ non-Real Complex zeros, which are Complex conjugates of one another.

Tschirnhaus transformation

To make the task of solving the cubic simpler, we make the cubic simpler using a linear substitution known as a Tschirnhaus transformation.

$0=27f(x)=27x^3-27x^2+189x+621$

$=(3x-1)^3+60(3x-1)+682$

$=t^3+60t+682$

where $t=(3x-1)$

Cardano's method

We want to solve:

$t^3+60t+682=0$

Let $t=u+v$ .

Then:

$u^3+v^3+3(uv+20)(u+v)+682=0$

Add the constraint $v=-20/u$ to eliminate the $(u+v)$ term and get:

$u^3-8000/u^3+682=0$

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

$(u^3)^2+682(u^3)-8000=0$

Use the quadratic formula to find:

$u^3=(-682+-sqrt((682)^2-4(1)(-8000)))/(2*1)$

$=(682+-sqrt(465124+32000))/2$

$=(682+-sqrt(497124))/2$

$=341+-3sqrt(13809)$

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

$t_1=root(3)(341+3sqrt(13809))+root(3)(341-3sqrt(13809))$

and related Complex roots:

$t_2=omega root(3)(341+3sqrt(13809))+omega^2 root(3)(341-3sqrt(13809))$

$t_3=omega^2 root(3)(341+3sqrt(13809))+omega root(3)(341-3sqrt(13809))$

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

Now $x=1/3(1+t)$ . So the roots of our original cubic are:

$x_1 = 1/3(1+root(3)(341+3sqrt(13809))+root(3)(341-3sqrt(13809)))$

$x_2 = 1/3(1+omega root(3)(341+3sqrt(13809))+omega^2 root(3)(341-3sqrt(13809)))$

$x_3 = 1/3(1+omega^2 root(3)(341+3sqrt(13809))+omega root(3)(341-3sqrt(13809)))$

Then the complete factorisation of our original quartic takes the form:

$x^4-2x^3+8x^2+16x-23 = (x-1)(x-x_1)(x-x_2)(x-x_3)$

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How do you list all possible roots and find all factors of $x^{4} + 2 x^{3} - 8 x^{2} + 16 x - 23$ $x^4+2x^3-8x^2+16x-23$ ?

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How do you list all possible roots and find all factors of $x^{4} + 2 x^{3} - 8 x^{2} + 16 x - 23$ $x^4+2x^3-8x^2+16x-23$ ?