Question

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

Answer 1

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

Explanation:

`f(x) = x^4-8x^3+7x-14`

By the rational root theorem, any rational zeros of `f(x)` are expressible in the form `p/q` for integers `p, q` with `p` a divisor of the constant term `-14` and `q` a divisor of the coefficient `1` of the leading term. So the only possible rational zeros are:

`+-1, +-2, +-7, +-14`

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

Can we solve it using other methods?

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

`x^4-8x^3+7x-14`

`=(x-2)^4-24(x-2)^2-57(x-2)-48`

`=t^4-24t^2-57t-48`

`=(t^2-at+b)(t^2+at+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.