easiest way to do Cubic factoring?

Question

Im stuck... I can't figure out how to do this, and since its the begining of the school year, theres no book to look into. So how do i do cubic factoring? as well as some harder quadratic factoring

y^3-3y^2+3y-1
z^3-9x^2+27z-27
u^3-2u^2-8u+16
25u^2-20uv+4v^2
27-t^3

these are the problems I'm stuck on

Thanks
Narumi

Answer 1

In short, there is no easy and general way.
However, for school work, since they don't teach and don't expect you to use the complicated formula to solve cupic equation, so, it means that the cupic expressions that you are expected to factorized have at least one 'nice' root, or the question itself provided some hint about or one of the root.

Let me try..

First one, the hint I guess is 1,3,3,1 being the coefficients, which is the coefficient if you expand
(a-b)^3. Try (y-1)^3 = y^3 -3y^2+3y-1. Lucky try.

Second one, (r u sure it is x^2, not z^2 in it?), see if some of the opposite signs can kill each others.
If z = 1, 27z-27 = 0 but not z^3-9z^2. How abuot z^3 and 27, hmm.. if z = 3, z^3-27 = 0, and -9z^2+27z = 0, thus (z-3) is a factor. Still quite lucky.

Third one, look at -8u + 16, try u = 2, -8u+16=0 and
u^3-2u^2 = 0! So lucky this time!

Fifth one, try t = 3, 27-t^3 = 0, so (t-3) is a factor.
27-t^3 = (t-3)(-t^2...), we have -t^3 and 3t^2, so
27-t^3 = (t-3)(-t^2-3t...), we have -t^3 and 0t^2 and 9t, so
27-t^3 = (t-3)(-t^2-3t-9), we have -t^3 and 0t^2 and 0t and 27. No remainder is expected since (t-3) is a factor.

Forth one, this one is different, it is not cubic, just a quadratic. The short cut is to notice it as a perfect square: (5u-2v)^2. If not, try factorizing v^2 out to have
v^2(25(u/v)^2 - 20(u/v) + 4), and treat u/v like single variable, say z. I am sure you can factorize 25z^2-20z+4.

Answer 2

here's some tips, (it's been a while for me so i'll just give you some reference) You can use Synthetic division to factor cubic polynomials. (see first link below)

Or if you have sum or difference of two cubes there is a short cut. (like your last example is 27-t^3, that is a difference of two cubes, because there are two cube numbers, 3^3 and t^3, and they are subtracted from one another)see the second link

Answer 3

The first one is just (y-1) cubed.
The second is (z-3) cubed.
The third is (u-2)(u^2-8), and that latter factors into (u+ sqr(8))(u- sqr(8))
The fourth is (5u - 2v) squared.
The fifth is (3-t)(9 + 3t + t^2)