BrowardMom wrote:How to factor 7n^5+7n^4-3n^2-6n-3 completely?

**To factor**, you use many of the same techniques that you would otherwise use to find the zeros, or solutions, of the equation "(polynomial) equal to (zero)".

The first step is usually to apply the

**Rational Roots Test**. This will give you a listing of

*possible* "nice" (fractional or whole-number) values to try.

Then you test these values by using

**synthetic division**.

In your case, the possible zeroes or roots are:

. . . . .Whole numbers are usually nicer, so try -1 and +1 first:

-1 | 7 7 0 -3 -6 -3

| -7 0 0 3 3

+----------------

7 0 0 -3 -3 0

Since this has a zero remainder, then n = -1 is a zero, so n + 1 is a factor.

Looking at the graph of the original polynomial, it appears that n = -1 is a zero twice, so:

-1 | 7 0 0 -3 -3

| -7 7 -7 10

+--------------

7 -7 7 -10 7

Okay, so there must be something else going on at n = -1; clearly, it is not a duplicated root.

If you try the various other potential roots, you'll discover that none works. So n = -1 is the only rational root, and n + 1 is the only "nice" factor. This leaves you with:

. . . . .I don't see anything that can be done with this, though...