y³ - 3y² - 4y + 12 

Factor the first two terms only:

y²(y - 3) - 4y + 12
 
Factor the last two terms only, taking out -4, remembering
that when you take a negative like -4 out of a positive like +12,
you get a negative (like -3).

y²(y - 3) - 4(y - 3)

I will color the like factors red

y²(y - 3) - 4(y - 3)

Thatke out the red factor (y - 3)

(y - 3)(y² - 4)

Now the second parenthetical expression will factor as the
difference of two squares

(y - 3)(y² - 2²)

(y - 3)(y - 2)(y + 2) 

====================================================

2x³ + 6x² - 8x - 24

First factor 2 out of the entire expression:

2[x³ + 3x² - 4x - 12]

Then factor the bracket as in the previous problem:

Factor x² out of first two terms in brackets:

2[x²(x + 3) - 4x - 12]

Factor -4 out of last two terms in brackets, remembering that
when you take a negative like -4 out of another negative, like -12,
you get a positive +3 :

2[x²(x + 3) - 4(x + 3)]

I'll color the like factors red:

2[x²(x + 3) - 4(x + 3)]

Take out the red factor from the terms inside the bracket:

2[(x + 3)(x² - 4)]

Now the second binomial will factor as the difference 
of two squares

2[(x + 3)(x² - 2²)]

2[(x + 3)(x - 2)(x + 2)]

Dispense with the brackets

2(x + 3)(x - 2)(x + 2) 

======================================================

6(2p+q)² - 5(2p+q) - 25 

6(2p+q)² - 5(2p+q) - 25

You can now treat the red parenthetical expression just as you
would treat a single letter.  That is, you can factor the above
just as you would factor 4x² - 5x - 25 as (3x + 5)(2x - 5), but
use brackets:

6(2p+q)² - 5(2p+q) - 25

[3(2p+q) + 5][2(2p+q) - 5]

Now remove the parentheses inside the brackets:

[6p + 3q + 5][4p + 2q - 5]

Change the brackets to parentheses:

(6p + 3q + 5)(4p + 2q - 5)

===============================================

25y2m - (x2n-2xn+1)

Write the first term as (5ym)2.

(5ym)2 - (x2n-2xn+1)

Write the second expression as [(xn)2-2(xn)+1]

(5ym)2 - [(xn)2-2(xn)+1]

(5ym)2 - [(xn)2-2(xn)+1]

Now factor the bracketed expression treating the red parentheses
as though they were just a single letter, but we'll need to go
to braces

(5ym)2 - {(xn)2-2(xn)+1}

(5ym)2 - {[(xn)-1][(xn)-1]}

Since the two bracketed factors in the braces are the same we 
can just write

(5ym)2 - (xn-1)2

(5ym)2 - (xn-1)2

Now to do some more coloring:

(5ym)2 - (xn-1)2

Now this is the difference of two squares and factors as

[(5ym) - (xn-1)][(5ym) + (xn-1)]

Remove the parentheses inside the bracket:

[5ym - xn + 1][5ym + xn - 1]

Change the brackets to parentheses:

(5ym - xn + 1)(5ym + xn - 1)

======================================================
I'm not going to color on the rest.  See if you can figure 
them out without colors:
======================================================

x6a  - t3b

Write these terms as

(x2a)3 - (tb)3

This is the difference of two cubes.  Use the rule for factoring
the sum or difference of two cubes:
                                                _
                           P³ + Q³ = (P + Q)(P² + PQ + Q²)

[(x2a) - (tb)][(x2a)2 + (x2a)(tb) + (tb)2] 

[x2a - tb][x4a + x2atb + t2b]

===============================================

(y-1)4 - (y-1) 

Factor out (y-1)

(y-1)[(y-1)³ - 1]

(y-1)[(y-1)³ - 1³]

The bracketed expression is the difference of two cubes, and so we
use the rule I gave in the last problem:

(y-1)[(y-1) - 1][(y-1)² + (y-1)(1) + 1²]

Remove all the parentheses inside the brackets:

(y-1)[y - 1 - 1][y²-2y+1 + y-1 + 1]

(y-1)[y-2}[y²-y+1]

(y-1)(y-2)(y²-y+1)

=============================================

X6 - 2X5 + X4 - X2 + 2X - 1

Factor X4 out of the first three terms:

X4(X2-2X+1) - X2 + 2X - 1

Factor -1 out of the last three terms, remembering to change
the sign when factoring out a negative like -1

X4(X2-2X+1) - 1(X2-2X+1)

Now factor out the common factor (X2-2X+1)

(X2-2X+1)[X4 - 1]

The first factor factors as (X-1)(X-1) = (X-1)2

(X-1)2[X4 - 1]

The second factor is the difference of two squares:

(X-1)2[(X2)2 - 12]

(X-1)2[(X2-1)(X2+1)]

The first factor in the brackets is the difference of two squares

(X-1)2[(X-1)(X+1)(X2+1)]

Dispense with the brackets

(X-1)2(X-1)(X+1)(X2+1)

The first two factors have the same base (X-1), so we can write them
as the cube of this:

(X-1)3(X+1)(X2+1)

Edwin
AnlytcPhil@aol.com