How would I Factor 21x^5-9x^3-7x^2+3?

Answer 1

You factor by grouping. Since addition is commutative, which means it doesn't matter what order the terms are in, you rearrange the terms so that you can make two groups that can then have common factors pulled from them.
Usually you start by dealing with the first half and then do the second half (two terms each) but since we can rearrange, we get a much better result with this arrangement:

21x^5-7x^2-9x^3+3
This makes two neat groups:
(21x^5 -7x^2) (-9x^3+3)
These each have common factors that can be pulled from them resulting in a repeating factor:
7x^2 (3x^3-1) -3(3x^3-1)
Now construct a factor out of the common factors you pulled: (7x^2-3) and you have one factor.
(3x^3-1) is the other.

Answer 2

-3x^3(-7x^2+3)-7x^2+3

(-7x^2+3)(-3x^3+1)