Question

How do you factor `r^3 - 2r^2 - 5r + 6 = 0`?

Answer 1

`(r-1)(r+2)(r-3)`

Explanation:

Using the rational root theorem, you can deduce that the possible roots of the function are `1, 2, 3, 6, -1, -2, -3,` and`-6` (which are simply possible factors of the term without `r`).

Now, just plug in some of the possible roots into the equation to see if it equals `0`. Plugging in `1`, we see that `1^3 -2(1^2)-5(1)+6` is `1-2-5+6` which equates to `0`.

Now that we have determined a root of the original equation, we can use synthetic division to divide the polynomial by the binomial `(x-1)`. Dividing, we get `r^2-r-6`.

Now it is simply factoring quadratic equation. When we factor, we get `(r+2)(r-3)`.

Combining these roots with our roots determined earlier, we can get the full answer as

`r^3-2r^2-5r+6=(r-1)(r+2)(r-3)`