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

1 Answer
Jan 28, 2016

Answer:

#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)#