How do you use the rational roots theorem to find all possible zeros of $f (x) = x^{4} - 8 x^{3} + 7 x - 14$ $f(x)=x^4-8x^3+7x-14$ ?

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How do you use the rational roots theorem to find all possible zeros of $f (x) = x^{4} - 8 x^{3} + 7 x - 14$ $f(x)=x^4-8x^3+7x-14$ ?

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Answer 1 · 2016-08-09T20:42:41

This $f (x)$ $f(x)$ has no rational zeros, but we can find the zeros algebraically...

Explanation:

$f (x) = x^{4} - 8 x^{3} + 7 x - 14$ $f(x) = x^4-8x^3+7x-14$

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 $- 14$ $-14$ and $q$ $q$ a divisor of the coefficient $1$ $1$ of the leading term. So the only possible rational zeros are:

$\pm 1, \pm 2, \pm 7, \pm 14$ $+-1, +-2, +-7, +-14$

None of these work, so $f (x)$ $f(x)$ has no rational zeros.

Can we solve it using other methods?

Substitute $t = (x - 2)$ $t = (x-2)$ to get a quartic with no cubed term...

$x^{4} - 8 x^{3} + 7 x - 14$ $x^4-8x^3+7x-14$

$= {(x - 2)}^{4} - 24 {(x - 2)}^{2} - 57 (x - 2) - 48$ $=(x-2)^4-24(x-2)^2-57(x-2)-48$

$= t^{4} - 24 t^{2} - 57 t - 48$ $=t^4-24t^2-57t-48$

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

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

Equating coefficients and rearranging slightly, we get:

${ (b+c=a^2-24), (b-c=-57/a), (bc=-48) :}$

Then:

$(a^2-24)^2 = (b+c)^2 = (b-c)^2+4bc = (-57/a)^2-192$

Expanding both ends, we get:

$(a^2)^2-48(a^2)+576 = 3249/((a^2))-192$

Add $192$ to both sides, multiply through by $(a^2)$ and rearrange slightly to get:

$0 = (a^2)^3-48(a^2)^2+768(a^2)-3249$

$= ((a^2)-16)^3+847$

Hence one Real root gives us:

$a^2 = 16-root(3)(847)$

which is positive.

So we can let $a=sqrt(16-root(3)(847))$

Substituting this into the set of simultaneous equations, we get:

${ (b+c=a^2-24=(16-root(3)(847))-24 = -8-root(3)(847)), (b - c = -57/a = -57/sqrt(16-root(3)(847))), (bc = -48) :}$

Add the first and second of these and divide by $2$ to find:

$b = -4 -1/2 root(3)(847) - 57/(2sqrt(16-root(3)(847)))$

Subtract the second from the first and divide by $2$ to find:

$c = -4 -1/2 root(3)(847) + 57/(2sqrt(16-root(3)(847)))$

So we now have two quadratics to solve:

$t^2-(sqrt(16-root(3)(847)))t-(4+1/2 root(3)(847)+57/(2sqrt(16-root(3)(847)))) = 0$

$t^2+(sqrt(16-root(3)(847)))t-(4+1/2 root(3)(847)-57/(2sqrt(16-root(3)(847)))) = 0$

We can solve these using the quadratic formula, then add $2$ to get zeros of our original quartic in $x$ .

I will leave that as an exercise.

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How do you use the rational roots theorem to find all possible zeros of $f (x) = x^{4} - 8 x^{3} + 7 x - 14$ $f(x)=x^4-8x^3+7x-14$ ?

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Explanation:

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How do you use the rational roots theorem to find all possible zeros of f(x)=x^4-8x^3+7x-14f(x)=x4−8x3+7x−14f(x)=x^4-8x^3+7x-14?

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How do you use the rational roots theorem to find all possible zeros of $f (x) = x^{4} - 8 x^{3} + 7 x - 14$ $f(x)=x^4-8x^3+7x-14$ ?