Question

How do you find the roots of `x^3-x^2-34x-56=0`?

Answer 1

This cubic equation has roots `7, -2, -4`.

Explanation:

`f(x) = x^3-x^2-34x-56`

By the rational roots 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 `-56` and `q` a divisor of the coefficient `1` of the leading term.

That means that the only possible rational zeros are:

`+-1, +-2, +-4, +-7, +-8, +-14, +-28, +-56`

In addition, note that the pattern of signs of the coefficients of the terms is `+ - - -`. By Descartes' Rule of Signs, since this has only one change of sign, `f(x)` has exactly one positive Real zero.

If `x >= 1` then `x^2+34x+56 >= 1+34+56 = 91`. So that positive Real zero must satisfy `x^3 >= 91`. So we can immediately rule out `1, 2, 4` as possibilities. Try `x = 7`:

`f(7) = 7^3-7^2-34(7)-56 = 343-49-238-56 = 0`

So `x=7` is a zero and `(x-7)` a factor:

`x^3-x^2-34x-56 = (x-7)(x^2+6x+8)`

We can factor the remaining quadratic by finding a pair of numbers with sum `6` and product `8`. The pair `2, 4` works, so we have:

`x^2+6x+8 = (x+2)(x+4)`

Putting it all together:

`x^3-x^2-34x-56 = (x-7)(x+2)(x+4)`

with zeros `7, -2, -4`