How do you use the rational roots theorem to find all possible zeros of $P (x) = 2 x^{4} + 10 x^{3} - 3 x^{2} - 8 x + 4$ $P(x) = 2x^4 + 10x^3 -3x^2 -8x +4$ ?

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How do you use the rational roots theorem to find all possible zeros of $P (x) = 2 x^{4} + 10 x^{3} - 3 x^{2} - 8 x + 4$ $P(x) = 2x^4 + 10x^3 -3x^2 -8x +4$ ?

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Answer 1 · 2016-08-20T20:41:31

We find there are no rational zeros, but it is possible to solve algebraically.

Explanation:

$P (x) = 2 x^{4} + 10 x^{3} - 3 x^{2} - 8 x + 4$ $P(x) = 2x^4+10x^3-3x^2-8x+4$

By the rational roots theorem, any rational zeros of $P (x)$ $P(x)$ 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 $4$ $4$ and $q$ $q$ a divisor of the coefficient $2$ $2$ of the leading term.

That means that the only possible rational zeros are:

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

Evaluating $P (x)$ $P(x)$ for each of these values, we find that none is a zero. So $P (x)$ $P(x)$ has no rational zeros.

In fact, it has two Real zeros and two non-Real Complex zeros. It is a little messy to solve algebraically, but it can be done...

Tschirnhaus transformation

First simplify the quartic using a linear substitution called a Tschirnhaus transformation...

$128 P (x) = 256 x^{4} + 1280 x^{3} - 384 x^{2} - 1024 x + 512$ $128P(x) = 256x^4+1280x^3-384x^2-1024x+512$

$= {(4 x + 5)}^{4} - 174 {(4 x + 5)}^{2} + 984 (4 x + 5) - 683$ $=(4x+5)^4-174(4x+5)^2+984(4x+5)-683$

$= t^{4} - 174 t^{2} + 984 t - 683$ $=t^4-174t^2+984t-683$

where $t = 4 x + 5$ $t = 4x+5$

Factorisation into quadratics

Since this quartic in $t$ $t$ is monic with no $t^{3}$ $t^3$ term, it will factor as the product of two monic quadratics with opposite middle terms:

$t^{4} - 174 t^{2} + 984 t - 683$ $t^4-174t^2+984t-683$

$= (t^{2} - a t + b) (t^{2} + a t + c)$ $=(t^2-at+b)(t^2+at+c)$

$= t^{4} + (b + c - a^{2}) t^{2} + (b - c) a t + b c$ $=t^4+(b+c-a^2)t^2+(b-c)at+bc$

Equating coefficients and rearranging slightly, we get:

${ (b+c=a^2-174), (b-c=984/a), (bc=-683) :}$

Hence:

$(a^2-174)^2 = (b+c)^2 = (b-c)^2+4bc = (984/a)^2+4(-683)$

Expanded:

$(a^2)^2-348(a^2)+30276 = 968256/((a^2))-2732$

Multiply through by $(a^2)$ and rearrange slightly to get:

$(a^2)^3-348(a^2)^2+33008(a^2)-968256 = 0$

We can solve this cubic in $a^2$ using Cardano's method to find Real root:

$a^2 = 4/3(87+2root(3)(-6885+60sqrt(1761))+2root(3)(-6885-60sqrt(1761))) ~~ 16.1274$

So we can let $a$ be the positive square root:

$a = sqrt(4/3(87+2root(3)(-6885+60sqrt(1761))+2root(3)(-6885-60sqrt(1761))))$

Then:

$b = 1/2(a^2-174+984/a)$

$c = 1/2(a^2-174-984/a)$

Hence two quadratics to solve to find $t$ and hence $x$ .

Messy, but not difficult.

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How do you use the rational roots theorem to find all possible zeros of $P (x) = 2 x^{4} + 10 x^{3} - 3 x^{2} - 8 x + 4$ $P(x) = 2x^4 + 10x^3 -3x^2 -8x +4$ ?

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How do you use the rational roots theorem to find all possible zeros of P(x) = 2x^4 + 10x^3 -3x^2 -8x +4P(x)=2x4+10x3−3x2−8x+4P(x) = 2x^4 + 10x^3 -3x^2 -8x +4?

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How do you use the rational roots theorem to find all possible zeros of $P (x) = 2 x^{4} + 10 x^{3} - 3 x^{2} - 8 x + 4$ $P(x) = 2x^4 + 10x^3 -3x^2 -8x +4$ ?