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
