Question

How do you factor `r^4 + r^3 − 3r^2 − 5r − 2`?

Answer 1

`r^4 + r^3 - 3r^2 - 5r - 2 = (r+1)^3(r-2)`

Explanation:

One way to do this is finding one root of the term and then performing polynomial long division. The procedure can be repeated until there is only a quadratic term left.

1) Searching for the first root / factor

If searching for a root, it is generally a good idea to evaluate the term for values like `r = 1`, `r = -1`, `r = 2`, `r = -2`, ...

Here, it works with `r = -1`:

`(-1)^4 + (-1)^3 - 3 (-1)^2 - 5 * (-1) - 2 = 1 - 1 - 3 + 5 - 2 = 0`

Thus, `r = -1` is a root and `(r - (-1))` is one of the factors of your term.

2) Polynomial long division

Let's use divide by `r + 1` to make the term easier and then hopefully find further roots.

`color(white)(xx)(r^4 + r^3 - 3r^2 - 5r - 2) -: (r + 1) = r^3 - 3r - 2`
`-(r^4 + r^3)`
`color(white)(x) color(white)(xxxxxx)/`
`color(white)(xxxxxx)0 - 3 r^2 - 5r`
`color(white)(xxxx)-( - 3 r^2 -3r)`
`color(white)(xxxxxx) color(white)(xxxxxxxxx)/`
`color(white)(xxxxxxxxxxx)-2r - 2`
`color(white)(xxxxxxxxx)-(-2r - 2)`
`color(white)(xxxxxxxxxxx) color(white)(xxxxxxxx)/`
`color(white)(xxxxxxxxxxxxxxxxx)0`

Thus, you already can factor your term as follows:

`r^4 + r^3 - 3r^2 - 5r - 2 = (r+1)(r^3 - 3r -2)`

3) Searching for the second root / factor

Now, let's try and factor `r^3 - 3r -2` further.

We can repeat the same procedure as before and again, we find that `r = -1` is a root:

`(-1)^3 - 3* (-1) - 2 = -1 + 3 - 2 = 0`

4) Polynomial long division

Thus, we can divide `(r^3 - 3r - 2)` by `(r + 1)` to simplify the term further:

`color(white)(xx) (r^3 color(white)(xxxx) - 3r - 2) -: (r+1) = r^2 - r - 2`
`- (r^3 + r^2)`
`color(white)(x) color(white)(xxxxxx) /`
`color(white)(xxxx) -r^2 - 3r`
`color(white)(xx) -(-r^2 - r)`
`color(white)(xxxx) color(white)(xxxxxxxx) /`
`color(white)(xxxxxxxx) -2r - 2`
`color(white)(xxxxxx) -(-2r - 2)`
`color(white)(xxxxxxxx) color(white)(xxxxxxxx) /`
`color(white)(xxxxxxxxxxxxxx) 0`

At this point, we can factor the term as follows:

`r^4 + r^3 - 3r^2 - 5r - 2 = (r+1)(r+1)(r^2 - r - 2)`

5) Factoring the quadratic term

At this point, the only thing left to do is factoring the term `r^2 - r - 2`.

There are a lot of ways to do that. Let me show you one of my favourites.

Basically, you would like to have something like this:

`r^2 - r - 2 = (r +a)(r+b)`

`= r^2 + (a + b)r + a * b`

Thus, you need to find `a` and `b` so that `a + b = -1` and `a times b = -2`

The solution to this is `a = 1` and `b = -2`.

Thus, your quadratic term can be factored as follows:

`r^2 - r - 2 = (r + 1)(r - 2)`

6) Solution

In total, you have found the following factorization:

`r^4 + r^3 - 3r^2 - 5r - 2 = (r+1)(r+1)(r+1)(r-2)`

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