Question

How do you use the rational root theorem to find the roots of `f(x)=x^3+x^2-8x-6`?

Answer 1

The zeros of `f(x)` are `-3`, `1+sqrt(3)` and `1-sqrt(3)`

Explanation:

Given:

`f(x) = x^3+x^2-8x-6`

The rational root theorem tells us that any rational zero of `f(x)` is expressible in the form `p/q` for integers `p, q` with `p` a divisor of the constant term `-6` and `q` a divisor of the coefficient `1` of the leading term.

That means that the only possible rational zeros are:

`+-1, +-2, +-3, +-6`

Trying each of these in turn, we eventually find:

`f(-3) = (color(blue)(-3))^3+(color(blue)(-3))^2-8(color(blue)(-3))-6 = -27+9+24-6 = 0`

So `x=-3` is a zero and `(x+3)` a factor:

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

We can find the zeros of the remaining quadratic by completing the square and using the difference of squares identity:

`A^2-B^2 = (A-B)(A+B)`

with`A=(x-1)` and `B=sqrt(3)` as follows:

`x^2-2x-2 = x^2-2x+1-3`

`color(white)(x^2-2x-2) = (x-1)^2-(sqrt(3))^2`

`color(white)(x^2-2x-2) = ((x-1)-sqrt(3))((x-1)+sqrt(3))`

`color(white)(x^2-2x-2) = (x-1-sqrt(3))(x-1+sqrt(3))`

Hence the other two zeros are:

`x = 1+-sqrt(3)`