Question

How do you use the rational root theorem to find the roots of `2x^3 + 7x^2 - 77x - 40`?

Answer 1

`{(x=-1/2), (x=-8), (x=5) :}`

Explanation:

`f(x) = 2x^3+7x^2-77x-40`

By the rational root theorem, any rational zeros of this polynomial must be expressible in the form `p/q` for integers `p`, `q` with `p` a divisor of the constant term `-40` and `qa` a divisor of the coefficient `2` of the leading term.

That means that the only possible rational zeros are:

`+-1/2`, `+-1`, `+-2`, `+-5/2`, `+-4`, `+-5`, `+-8`, `+-10`, `+-20`, `+-40`

This is rather a lot of possibilities to try, but trying each in turn we soon find:

`f(-1/2) = -1/4+7/4+77/2-40 = (-1+7+154-160)/4 = 0`

So `x=-1/2` is a zero and `(2x+1)` a factor:

`2x^3+7x^2-77x-40 = (2x+1)(x^2+3x-40)`

To factor the remaining quadaratic find a pair of factors of `40` that differ by `3`. The pair `8, 5` works.

Hence:

`x^2+3x-40 = (x+8)(x-5)`

So the other two zeros are `x=-8` and `x=5`