Question 24971
<pre><b><font face = "lucida console" size = 4>
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<font color = "red">(y - 3)</font> - 4<font color = "red">(y - 3)</font>

Thatke out the red factor <font color = "red">(y - 3)</font>

<font color = "red">(y - 3)</font>(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<font color = "red">(x + 3)</font> - 4<font color = "red">(x + 3)</font>]

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

2[<font color = "red">(x + 3)</font>(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<font color = "red">(2p+q)</font>² - 5<font color = "red">(2p+q)</font> - 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 4<font color = "red">x</font>² - 5<font color = "red">x</font> - 25 as (3<font color = "red">x</font> + 5)(2<font color = "red">x</font> - 5), but
use brackets:

6<font color = "red">(2p+q)</font>² - 5<font color = "red">(2p+q)</font> - 25

[3<font color = "red">(2p+q)</font> + 5][2<font color = "red">(2p+q)</font> - 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)

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

25y<sup>2m</sup> - (x<sup>2n</sup>-2x<sup>n</sup>+1)

Write the first term as (5y<sup>m</sup>)<sup>2</sup>.

(5y<sup>m</sup>)<sup>2</sup> - (x<sup>2n</sup>-2x<sup>n</sup>+1)

Write the second expression as [(x<sup>n</sup>)2-2(x<sup>n</sup>)+1]

(5y<sup>m</sup>)<sup>2</sup> - [(x<sup>n</sup>)<sup>2</sup>-2(x<sup>n</sup>)+1]

(5y<sup>m</sup>)<sup>2</sup> - [<font color = "red">(x<sup>n</sup>)</font>2-2<font color = "red">(x<sup>n</sup>)</font>+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

(5y<sup>m</sup>)<sup>2</sup> - {<font color = "red">(x<sup>n</sup>)</font>2-2<font color = "red">(x<sup>n</sup>)</font>+1}

(5y<sup>m</sup>)<sup>2</sup> - {[<font color = "red">(x<sup>n</sup>)</font>-1][<font color = "red">(x<sup>n</sup>)</font>-1]}

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

(5y<sup>m</sup>)<sup>2</sup> - (<font color = "red">x<sup>n</sup></font>-1)2

(5y<sup>m</sup>)<sup>2</sup> - (x<sup>n</sup>-1)<sup>2</sup>

Now to do some more coloring:

<font color = "blue">(5y<sup>m</sup>)</font><sup>2</sup> - <font color = "blue">(x<sup>n</sup>-1)</font><sup>2</sup>

Now this is the difference of two squares and factors as

[<font color = "blue">(5y<sup>m</sup>)</font> - <font color = "blue">(x<sup>n</sup>-1)</font>][<font color = "blue">(5y<sup>m</sup>)</font> + <font color = "blue">(x<sup>n</sup>-1)</font>]

Remove the parentheses inside the bracket:

[5y<sup>m</sup> - x<sup>n</sup> + 1][5y<sup>m</sup> + x<sup>n</sup> - 1]

Change the brackets to parentheses:

(5y<sup>m</sup> - x<sup>n</sup> + 1)(5y<sup>m</sup> + x<sup>n</sup> - 1)

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

x<sup>6a</sup>  - t<sup>3b</sup>

Write these terms as

(x<sup>2a</sup>)<sup>3</sup> - (t<sup>b</sup>)<sup>3</sup>

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

[(x<sup>2a</sup>) - (t<sup>b</sup>)][(x<sup>2a</sup>)<sup>2</sup> + (x<sup>2a</sup>)(t<sup>b</sup>) + (t<sup>b</sup>)<sup>2</sup>] 

[x<sup>2a</sup> - t<sup>b</sup>][x<sup>4a</sup> + x<sup>2a</sup>t<sup>b</sup> + t<sup>2b</sup>]

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

(y-1)<sup>4</sup> - (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)

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

X<sup>6</sup> - 2X<sup>5</sup> + X<sup>4</sup> - X<sup>2</sup> + 2X - 1

Factor X<sup>4</sup> out of the first three terms:

X<sup>4</sup>(X<sup>2</sup>-2X+1) - X<sup>2</sup> + 2X - 1

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

X<sup>4</sup>(X<sup>2</sup>-2X+1) - 1(X<sup>2</sup>-2X+1)

Now factor out the common factor (X<sup>2</sup>-2X+1)

(X<sup>2</sup>-2X+1)[X<sup>4</sup> - 1]

The first factor factors as (X-1)(X-1) = (X-1)<sup>2</sup>

(X-1)<sup>2</sup>[X<sup>4</sup> - 1]

The second factor is the difference of two squares:

(X-1)<sup>2</sup>[(X<sup>2</sup>)<sup>2</sup> - 1<sup>2</sup>]

(X-1)<sup>2</sup>[(X<sup>2</sup>-1)(X<sup>2</sup>+1)]

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

(X-1)<sup>2</sup>[(X-1)(X+1)(X<sup>2</sup>+1)]

Dispense with the brackets

(X-1)<sup>2</sup>(X-1)(X+1)(X<sup>2</sup>+1)

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

(X-1)<sup>3</sup>(X+1)(X<sup>2</sup>+1)

Edwin
AnlytcPhil@aol.com</pre>