Question

How do you use the rational root theorem to find the roots of `f(x) = 9x^8+9x^6-12x+7`?

Answer 1

Use the Durand-Kerner method to find approximations:

`x = 0.676693+-0.0518119i`

`x = 0.456213+-0.953409i`

`x = -0.902683+-0.483723i`

`x = -0.230223+-1.17823i`

Explanation:

`f(x) = 9x^8+9x^6-12x+7`

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 `7` and `q` a divisor of the coefficient `9` of the leading term.

So the only possible rational zeros are:

`+-1/9`, `+-1/3`, `+-7/9`, `+-1`, `+-7/3`, `+-7`

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

We can find approximations to the zeros using the Durand-Kerner method.

Suppose the zeros are `p, q, r, s, t, u, v, w`.

Choose initial approximations:

`p_0 = (0.4+0.9i)^0`

`q_0 = (0.4+0.9i)^1`

`r_0 = (0.4+0.9i)^2`
`vdots`
`w_0 = (0.4+0.9i)^7`

Iterate using the formulas:

`p_(k+1) = p_k - f(p_k)/(9(p_k - q_k)(p_k - r_k)...(p_k-w_k))`

`q_(k+1) = q_k - f(q_k)/(9(q_k - p_(k+1))(q_k - r_k)...(q_k-w_k))`

`r_(k+1) = r_k - f(r_k)/(9(r_k - p_(k+1))(r_k - q_(k+1))...(r_k-w_k))`
`vdots`
`w_(k+1) = w_k - f(w_k)/(9(w_k - p_(k+1))(w_k - q_(k+1))...(w_k-v_(k+1)))`

Note the factor `9` in the denominator is required to effectively make `f(x)` into a monic polynomial to get the Durand-Kerner method to work.

It only takes about `8` iterations to get to the approximations:

`x = 0.676693+-0.0518119i`

`x = 0.456213+-0.953409i`

`x = -0.902683+-0.483723i`

`x = -0.230223+-1.17823i`