Question

Question #d4732

Answer 1

`x=3,x~~-2.81`

Explanation:

We begin by moving everything over to one side so we are looking for zeroes of a polynomial:
`x^6-x^2-40x-600=0`

We can now use the Rational Roots Theorem to find that the possible rational zeroes are all the coefficients of `600` (the first coefficient is `1`, and dividing by `1` doesn't make a difference).

This gives the following rather large list:
`+-1,+-2,+-3,+-4,+-5,+-6,+-8,+-10,+-12,+-15,+-20,+-24,+-25,+-30,+-40,+-50,+-60,+-75,+-100,+-120,+-150,+-200,+-300,+-600`

Luckily, we quite quickly get that `x=3` is a zero. This means that `x=3` is a solution to the original equation.

There is a negative solution to this equation as well, but it is not rational, so we cannot find it using the Rational Roots Theorem.

Using Polynomial Long Division and the fact that `(x-3)` will be a factor only helps us reduce the equation to a degree five equation, which we still can't solve.

Our only remaining option is to use one of the available approximation methods. Using Newton's method, we get that there is a solution around `x~~-2.81`.